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;
long long n, m, mod, l, r, u, v;
long long sum(long long gl, long long gr, long long pl, long long pr,
              long long step) {
  if (gl > r) return 0;
  if (u <= pl && v >= pr) {
    if (gr < gl) return 0;
    long long old_gl = gl;
    gl = ((l - gl) / step) * step + gl;
    if (gl < l) gl += step;
    gr = ((r - old_gl) / step) * step + old_gl;
    if (gr < gl)
      return 0;
    else {
      long long n = (gr - gl) / step + 1, z = gl + gr;
      if (!(n % 2))
        return (((n / 2) % mod) * (z % mod) + n) % mod;
      else
        return (((z / 2) % mod) * (n % mod) + n) % mod;
    }
  }
  if (gl + step >= gr) {
    if (gl >= l && gl <= r) {
      return (gl + 1) % mod;
    } else {
      return 0;
    }
  }
  long long pm = (pl + pr + 1) / 2, total = 0;
  if (u < pm) total = (total + sum(gl, gr, pl, pm, step * 2)) % mod;
  if (v >= pm)
    total = (total + sum(gl + step, gr + step * (((pr - pl) % 2 == 0) ? 1 : -1),
                         pm, pr, step * 2)) %
            mod;
  return total;
}
int main() {
  scanf("%I64d %I64d %I64d", &n, &m, &mod);
  for (int i = 0; i < m; i++) {
    scanf("%I64d %I64d %I64d %I64d", &u, &v, &l, &r);
    u--;
    v--;
    l--;
    r--;
    if (v > n - 1) v = n - 1;
    if (r > n - 1) r = n - 1;
    if (r < l || v < u)
      cout << 0 << endl;
    else
      cout << sum(0, n, 0, n, 1) << endl;
  }
}