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;
int rst;
long long u, v;
void Query(long long L, long long R, long long l, long long r, long long e,
           int x) {
  long long r1, r2;
  if (L == l && R == r) {
    if (!x) {
      r1 = v + 1;
      r2 = v + 2;
      if (r1 % 2)
        r2 /= 2;
      else
        r1 /= 2;
      r1 %= M, r2 %= M;
      rst = (rst + (r1 * r2) % M) % M;
      r1 = u, r2 = u + 1;
      if (r1 & 1)
        r2 /= 2;
      else
        r1 /= 2;
      r1 %= M;
      r2 %= M;
      rst = (rst + M - (r1 * r2) % M) % M;
    } else {
      long long le = ((u >> x) << x) + e, ri = ((v >> x) << x) + e;
      if (le < u) le += 1LL << x;
      if (ri > v) ri -= 1LL << x;
      if (le <= ri) {
        r1 = le + ri + 2;
        r2 = (ri >> x) - (le >> x) + 1;
        if (r1 % 2)
          r2 >>= 1;
        else
          r1 >>= 1;
        r1 %= M, r2 %= M;
        rst = (rst + r1 * r2) % M;
      }
    }
    return;
  }
  long long m = (L + R) >> 1;
  if (l <= m) Query(L, m, l, min(r, m), e, x + 1);
  if (m < r) Query(m + 1, R, max(m + 1, l), r, e + (1LL << x), x + 1);
}
int main() {
  long long n;
  int m;
  scanf("%I64d%d%d", &n, &m, &M);
  while (m--) {
    long long l, r;
    scanf("%I64d%I64d%I64d%I64d", &l, &r, &u, &v);
    if (u > n) {
      puts("0");
      continue;
    }
    if (v > n) v = n;
    rst = 0;
    u--;
    v--;
    Query(0, n - 1, l - 1, r - 1, 0, 0);
    printf("%d\n", rst);
  }
  return 0;
}