Problem: Let a be an array consisting of n numbers. The array's elements are numbered from 1 to n, even is an array consisting of the numerals whose numbers are even in a (eveni = a2i, 1 ≤ 2i ≤ n), odd is an array consisting of the numberals whose numbers are odd in а (oddi = a2i - 1, 1 ≤ 2i - 1 ≤ n). Then let's define the transformation of array F(a) in the following manner:

  * if n > 1, F(a) = F(odd) + F(even), where operation " + " stands for the arrays' concatenation (joining together) 
  * if n = 1, F(a) = a



Let a be an array consisting of n numbers 1, 2, 3, ..., n. Then b is the result of applying the transformation to the array a (so b = F(a)). You are given m queries (l, r, u, v). Your task is to find for each query the sum of numbers bi, such that l ≤ i ≤ r and u ≤ bi ≤ v. You should print the query results modulo mod.

Input

The first line contains three integers n, m, mod (1 ≤ n ≤ 1018, 1 ≤ m ≤ 105, 1 ≤ mod ≤ 109). Next m lines describe the queries. Each query is defined by four integers l, r, u, v (1 ≤ l ≤ r ≤ n, 1 ≤ u ≤ v ≤ 1018).

Please do not use the %lld specificator to read or write 64-bit integers in C++. Use %I64d specificator.

Output

Print m lines each containing an integer — remainder modulo mod of the query result.

Examples

Input

4 5 10000
2 3 4 5
2 4 1 3
1 2 2 4
2 3 3 5
1 3 3 4


Output

0
5
3
3
3


Input

2 5 10000
1 2 2 2
1 1 4 5
1 1 2 5
1 1 1 3
1 2 5 5


Output

2
0
0
1
0

Note

Let's consider the first example. First let's construct an array b = F(a) = F([1, 2, 3, 4]). 

  * Step 1. F([1, 2, 3, 4]) = F([1, 3]) + F([2, 4])
  * Step 2. F([1, 3]) = F([1]) + F([3]) = [1] + [3] = [1, 3]
  * Step 3. F([2, 4]) = F([2]) + F([4]) = [2] + [4] = [2, 4]
  * Step 4. b = F([1, 2, 3, 4]) = F([1, 3]) + F([2, 4]) = [1, 3] + [2, 4] = [1, 3, 2, 4]

Thus b = [1, 3, 2, 4]. Let's consider the first query l = 2, r = 3, u = 4, v = 5. The second and third positions in the array b do not have numbers in the range [4, 5], so the sum obviously equals zero. Let's consider the second query l = 2, r = 4, u = 1, v = 3. The second and third positions have two numbers that belong to the range [1, 3], their sum equals 5.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int m;
long long n, mod;
long long Sum(long long val) {
  if (val % 2) {
    return (val + 1) / 2 % mod * (val % mod) % mod;
  }
  return val / 2 % mod * ((val + 1) % mod) % mod;
}
long long arifm(long long len, long long f, long long step, long long l,
                long long r) {
  long long v1, v2;
  if (l < f) {
    v1 = 0;
  } else {
    v1 = (l - f) / step;
    while (f + v1 * step < l) {
      v1++;
    }
  }
  if (r > f + len * step) {
    v2 = len;
  } else {
    v2 = (r - f) / step;
  }
  len = v2 - v1 + 1;
  long long rez = (f % mod) * (len % mod) % mod;
  long long sum = ((Sum(v2) - Sum(v1 - 1) + mod) * (step % mod) + rez) % mod;
  return sum;
}
long long count(long long l, long long r, long long d, long long step,
                long long ll, long long rr, long long u, long long v) {
  if (r < ll || rr < l) {
    return 0;
  }
  if (v < d || d + step * (r - l) < u) {
    return 0;
  }
  if (ll <= l && r <= rr) {
    return arifm(r - l, d, step, u, v);
  }
  long long m = (l + r) / 2;
  return (count(l, m, d, step * 2, ll, rr, u, v) +
          count(m + 1, r, d + step, step * 2, ll, rr, u, v)) %
         mod;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m >> mod;
  for (; m; m--) {
    long long l, r, u, v;
    cin >> l >> r >> u >> v;
    long long ans = count(1, n, 1, 1, l, r, u, v);
    cout << ans << endl;
  }
  return 0;
}