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, u, v, l, r, pow2[70];
int m, bigp, ans;
void search(long long nownum, int depth, long long l, long long r,
            long long left, long long right) {
  long long mid = (left + right) / 2;
  if ((left == l) && (right == r)) {
    long long u1 = u - nownum, v1 = v - nownum;
    if (u1 < 0) u1 = 0;
    if (v1 < u1) return;
    if (u1 != 0) u1 = (u1 - 1) / pow2[depth] + 1;
    v1 = v1 / pow2[depth];
    long long c1 = u1 + v1, c2 = v1 - u1 + 1;
    if (c1 % 2 == 0)
      c1 /= 2;
    else
      c2 /= 2;
    c1 %= bigp, c2 %= bigp;
    ans = ((long long)ans +
           ((long long)c1 * c2 % bigp) * (long long)(pow2[depth] % bigp) +
           (long long)(v1 - u1 + 1) * (nownum + 1) % bigp) %
          bigp;
  } else {
    if (l <= mid) search(nownum, depth + 1, l, min(r, mid), left, mid);
    if (r >= mid + 1)
      search(nownum + pow2[depth], depth + 1, max(l, mid + 1), r, mid + 1,
             right);
  }
}
int main() {
  pow2[0] = 1;
  for (int i = 1; i <= 62; i++) pow2[i] = pow2[i - 1] * 2;
  scanf("%I64d %d %d", &n, &m, &bigp);
  for (int i = 1; i <= m; i++) {
    scanf("%I64d %I64d %I64d %I64d", &l, &r, &u, &v);
    if (u > n)
      ans = 0;
    else {
      if (v > n) v = n;
      --u, --v;
      ans = 0;
      search(0, 0, l, r, 1, n);
    }
    printf("%d\n", ans);
  }
  return 0;
}