For this Question: There are N robots standing in a row. Ultron wants to delegate a task of difficulty X to two of them. Each robot has its own ability, A[i] for the i<sup>th</sup> robot from the left. The task can only be completed if the sum of the abilities of the two robots is exactly equal to X. Ultron wants to find out the number of distinct unordered pairs of robots he can choose to complete this task. He wants to find this number for Q (not necessarily distinct) subarrays of robots. 

Formally, for each query, you will be given two integers L and R and you need to calculate the number of pairs (i, j) such that L ≤ i < j ≤ R and A[i] + A[j] = X.

Further, these subarrays have a nice property that for every pair of them at least one of the following three statements hold: 

1. Their intersection is one of the two intervals itself (for example, segments [2, 6] and [2, 4]).
2. Their intersection is empty (for example, segments [2, 6] and [7, 8]).
3. Their intersection consists of exactly one element (for example, segments [2, 5] and [5, 8]).

For another example, note that the segments [2, 5] and [4, 7] do not satisfy any of the above conditions, and thus cannot both appear in the input. 

You need to find the answer to each one of Ultron's queries and report the XOR of their answers.

<b>Note:</b> Since the input is large, it is recommended that you use fast input/output methods.The first line of the input contains two space-separated integers, N, the number of robots and X, the difficulty of the task.
The second line contains N space-separated integers  A[1], A[2], ... A[n] - the abilities of the robots.
The third line of input contains an integer Q - the number of queries Ultron has.
Q lines follow, the i<sup>th</sup> of them contains two integers L and R for the i<sup>th</sup> query.
It is guaranteed that the given subarrays satisfy the properties mentioned in the problem statement.

<b>Constraints: </b>
2 ≤ N ≤ 400000
1 ≤ X ≤ 1000000
0 ≤ A[i] ≤ X
1 ≤ Q ≤ 400000
1 ≤ L < R ≤ N for each queryPrint a single integer, the bitwise XOR of the answers of the Q queries.Sample Input:
7 10
2 9 1 8 4 4 6
5
1 7
2 5
1 6
3 4
4 5

Sample Output:
7

Explanation:

For the first query there are 4 pairs of indices, (2, 3), (1, 4), (5, 7), (6, 7).
For the second query there is 1 pair of indices, (2, 3)
Similarly you can verify that the answers for the five queries are (4, 1, 2, 0, 0) in order. So we print their bitwise XOR = 7. 
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;

#define int long long

int X;
vector<int> adj[400005];
int ans[400005];
int cnt[1000005];
int A[400005];
pair<int, int> Q[400005];
void dfs(int x, bool keep = false)
{
    int bigChild = -1, sz = 0;
    for(auto &p: adj[x])
    {
        if(Q[p-1].second - Q[p-1].first > sz)
        {
            sz = Q[p-1].second - Q[p-1].first;
            bigChild = p;
        }
    }

    if(x > 0 && bigChild == -1)
    {
        for(int cur = Q[x-1].first; cur <= Q[x-1].second; cur++)
        {
                ans[x-1] += cnt[X-A[cur]]; 
                cnt[A[cur]]++;
        }
        if(!keep)
        {
            for(int cur = Q[x-1].first; cur <= Q[x-1].second; cur++)
            {
                cnt[A[cur]]--;
            }
        }   
        return;
    }

    for(auto &p: adj[x])
    {
        if(p != bigChild)
        {
            dfs(p);
        }
    }

    dfs(bigChild, true);
    
    if(x > 0)
    {
        ans[x-1] = ans[bigChild-1];
        for(int cur = Q[x-1].first; cur <= Q[x-1].second; cur++)
        {
            if(cur == Q[bigChild-1].first)
            {
                cur = Q[bigChild-1].second;
                continue;
            }
            ans[x-1] += cnt[X-A[cur]];
            cnt[A[cur]]++;
        }
        if(!keep)
        {
            for(int cur = Q[x-1].first; cur <= Q[x-1].second; cur++)
            {
                cnt[A[cur]]--;
            }
        }  
    }
}


bool my(int a, int b)
{
    if(Q[a].first < Q[b].first) return true;
    else if(Q[a].first == Q[b].first) return Q[a].second > Q[b].second;
    return false;
}
void sack(int n, int q)
{
    vector<int> order;
    for(int i=0; i<q; i++) order.push_back(i);
    sort(order.begin(), order.end(), my);

    stack<int> right_ends;
    stack<int> st;
    for(int i=0; i<q; i++)
    {
        while(right_ends.size() && right_ends.top() <= Q[order[i]].first)
        {
            int which = st.top();
            st.pop();
            right_ends.pop();
            if(st.empty())
            {
                adj[0].push_back(which+1);
            }
            else
            {
                adj[st.top()+1].push_back(which+1);
            }
        }
        st.push(order[i]);
        // cout << "inserting " << queries[i][2] << " into stack\n";
        right_ends.push(Q[order[i]].second);
    }
    while(!st.empty())
    {
        int which = st.top();
        st.pop();
        right_ends.pop();
        if(st.empty())
        {
            adj[0].push_back(which+1);
        }
        else
        {
            adj[st.top()+1].push_back(which+1);
        }
    }

    // for(int i=0; i<=q; i++)
    // {
    //     for(auto &x: adj[i])
    //     {
    //         cout << i << " has child " << x << "\n";
    //     } 
    // }

    dfs(0);
}

signed main()
{
    // freopen("input.txt", "r", stdin);
    // freopen("output_sack.txt", "w", stdout);
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n;
    cin >> n >> X;
    for(int i=0; i<n; i++)
    {
        cin >> A[i];
    }
    int q;
    cin >> q;
    for(int i=0; i<q; i++)
    {
        cin >> Q[i].first >> Q[i].second;
        Q[i].first--;
        Q[i].second--;
    }

    
    sack(n, q);

    int ansr = 0;
    for(int i=0; i<q; i++)
    {
        ansr ^= ans[i];
    }
    cout << ansr << "\n";
    return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not?