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 mod;
long long Div(long long x, long long y) {
  return x < 0 ? (x - y + 1) / y : x / y;
}
long long sum(long long first, long long step, long long cl, long long cr,
              long long l, long long r, long long u, long long v) {
  if (cl >= l && cr <= r) {
    long long ut = max(0ll, u - first);
    long long t1 = max(0ll, (ut / step + (ut % step ? 1 : 0)));
    long long vt = v - first;
    long long t2 = min(Div(vt, step), cr - cl);
    long long num = t2 - t1 + 1;
    if (num >= 1) {
      long long ff = (first + step * t1) % mod;
      if (num == 1) return ff;
      long long x = (num - 1);
      if ((num - 1) % 2 == 0)
        x = (x / 2) % mod * (num % mod) % mod;
      else
        x = x % mod * (num / 2 % mod) % mod;
      return (ff * (num % mod) % mod + x * (step % mod) % mod) % mod;
    } else
      return 0;
  }
  long long num = (cr - cl + 1) / 2 + (cr - cl + 1) % 2;
  long long rt = 0;
  if (cl + num - 1 >= l)
    rt += sum(first, step * 2, cl, cl + num - 1, l, r, u, v);
  if (cl + num - 1 < r)
    rt += sum(first + step, step * 2, cl + num, cr, l, r, u, v);
  return rt % mod;
}
int main() {
  long long n;
  int m;
  scanf("%I64d%d%I64d", &n, &m, &mod);
  for (int i = 0; i < (m); i++) {
    long long l, r, u, v;
    scanf("%I64d%I64d%I64d%I64d", &l, &r, &u, &v);
    long long rt = sum(1, 1, 1, n, l, r, u, v) % mod;
    printf("%d\n", (int)rt);
  }
  return 0;
}