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;
template <class A, class B>
A convert(B x) {
  stringstream ss;
  ss << x;
  A ret;
  ss >> ret;
  return ret;
}
const int oo = ~0u >> 2;
const double eps = 1e-10;
const int fx[8][2] = {{-1, 0},  {1, 0},  {0, -1}, {0, 1},
                      {-1, -1}, {-1, 1}, {1, -1}, {1, 1}};
long long n, m, u, v, mo;
long long dg(long long a, long long b, long long n, long long l, long long r) {
  if (l == 0 && r == n) {
    long long x = min(n, max(0LL, (u - b + a - 1) / a));
    long long y = min(n, max(0LL, (v - b + a - 1) / a));
    if (x < y) {
      long long ret = 0;
      if ((y - x) % 2 == 0)
        ret = ((y - x) / 2 % mo) * ((x + y - 1) % mo);
      else
        ret = ((x + y - 1) / 2 % mo) * ((y - x) % mo);
      ret = ret % mo * (a % mo) + (y - x) % mo * (b % mo);
      return ret % mo;
    } else
      return 0LL;
  } else {
    long long m = (n + 1) / 2;
    if (r <= m) return dg(2 * a, b, m, l, r);
    if (m <= l) return dg(2 * a, b + a, n - m, l - m, r - m);
    return (dg(2 * a, b, m, l, m) + dg(2 * a, b + a, n - m, 0, r - m)) % mo;
  }
}
int main() {
  scanf("%I64d%I64d%I64d", &n, &m, &mo);
  while (m--) {
    long long l, r;
    scanf("%I64d%I64d%I64d%I64d", &l, &r, &u, &v);
    ++v;
    printf("%I64d\n", dg(1, 1, n, l - 1, r));
  }
}