For this Question: Given a range from l to r, Your task is to find the sum of the second last digit of the prime numbers between that range.The first line of the input contains the T test cases.
Next T lines contains the range value l and r.

<b>Constraints</b>
1 <= T <= 100
1 <= l < r <= 1e5
Difference between l and r dosen't exceed 1e4.Print the required sum.Sample Input 1:
3
2 15
1 20
3 10

Sample Output 1:
2
4
0, I have written this Solution Code: #include <bits/stdc++.h>

using namespace std;

#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif

vector<bool> sieve(int n) {
    vector<bool> is_prime(n + 1, true);
    is_prime[0] = is_prime[1] = false;
    for (int i = 2; i * i <= n; i++) {
        if (is_prime[i]) {
            for (int j = i * i; j <= n; j += i)
                is_prime[j] = false;
        }
    }
    return is_prime;
}

int main() {
    vector<bool> prime = sieve(1e5 + 1);
    int tt;
    cin >> tt;
    while (tt--) {
        int l, r, sum = 0;
        cin >> l >> r;
        for (int i = l; i <= r; i++) {
            if (prime[i]) {
                sum += ((i % 100) / 10);
                debug(i, (i % 100) / 10);
            }
        }
        cout << sum << "\n";
    }
    return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not?