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, L, R, U, V;
int M;
int mod;
int calc(long long a1, long long d, long long &le, long long &ri) {
  long long an = a1 + (ri - le) * d;
  if (U > an || V < a1) return 0;
  long long kr = (V - a1) / d;
  if (kr > ri - le) kr = ri - le;
  long long kl = (U <= a1 ? 0 : (U - 1 - a1) / d + 1);
  long long sum = a1 + kl * d + a1 + kr * d;
  long long g = kr - kl + 1;
  (sum & 1) ? g >>= 1 : sum >>= 1;
  return (g % mod) * (sum % mod) % mod;
}
long long query(long long a1, long long d, long long le, long long ri) {
  if (le > R || ri < L) return 0;
  if (L <= le && ri <= R) return calc(a1, d, le, ri);
  long long mid = (le + ri) >> 1;
  long long p = query(a1, d << 1, le, mid);
  long long q = query(a1 + d, d << 1, mid + 1, ri);
  return p + q;
}
int main() {
  for (; scanf("%lld%d%d", &N, &M, &mod) != EOF;) {
    for (int i = (1), _i = (M); i <= _i; ++i) {
      scanf("%lld%lld%lld%lld", &L, &R, &U, &V);
      int ans = query(1, 1, 1, N) % mod;
      printf("%d\n", ans);
    }
  }
  return 0;
}