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>
int md;
int s(long long n) {
  return (n % 2 == 0 ? (n / 2 % md) * ((n + 1) % md)
                     : (n % md) * ((n + 1) / 2 % md)) %
         md;
}
void solve(long long n, long long k, long long a, int *sum, int *cnt) {
  int sum0, cnt0, sum1, cnt1;
  if (k <= 0 || a <= 0)
    *sum = *cnt = 0;
  else if (k >= n) {
    if (a > n) a = n;
    *sum = s(a), *cnt = a % md;
  } else {
    solve((n + 1) / 2, k, (a + 1) / 2, &sum0, &cnt0);
    solve(n / 2, k - (n + 1) / 2, a / 2, &sum1, &cnt1);
    *sum = ((long long)sum0 * 2 - cnt0 + md + sum1 * 2) % md;
    *cnt = (cnt0 + cnt1) % md;
  }
}
int main() {
  long long n;
  int m;
  scanf("%lld%d%d", &n, &m, &md);
  while (m--) {
    long long l, r, a, b;
    int sum, cnt, ans;
    scanf("%lld%lld%lld%lld", &l, &r, &a, &b), l--, a--;
    ans = 0;
    solve(n, r, b, &sum, &cnt), ans = (ans + sum) % md;
    solve(n, r, a, &sum, &cnt), ans = (ans - sum + md) % md;
    solve(n, l, b, &sum, &cnt), ans = (ans - sum + md) % md;
    solve(n, l, a, &sum, &cnt), ans = (ans + sum) % md;
    printf("%d\n", ans);
  }
  return 0;
}