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, m, p, a, b, h;
long long rev(long long n) {
  int i;
  long long r;
  for (r = 0, i = 0; i < p; r = r * 2 + (n & 1), n >>= 1, i++)
    ;
  return r;
}
long long sm(long long f, long long d) {
  long long x = a, y = b;
  if (y < f) return 0;
  if (x < f) x = f;
  if ((x - f) % d) x += d - (x - f) % d;
  if ((y - f) % d) y -= (y - f) % d;
  if (x > y) return 0;
  if ((x + y) % 2 == 0)
    return ((x + y) / 2 % m * (((y - x) / d + 1) % m) + (y - x) / d + 1) % m;
  else
    return (((y - x) / d + 1) / 2 % m * ((x + y) % m) + (y - x) / d + 1) % m;
}
long long gt(long long i, long long j, long long l, long long r) {
  if (i > r || j < l) return 0;
  if (l >= i && r <= j) return sm(rev(l), h / (r - l + 1));
  return (gt(i, j, l, (l + r) / 2) + gt(i, j, (l + r) / 2 + 1, r)) % m;
}
long long pos(long long a, long long n, long long d) {
  if (n == 1) return 0;
  if (a >= (n + 1) / 2)
    return pos(a - (n + 1) / 2, n - (n + 1) / 2, 2 * d) + d;
  else
    return pos(a, (n + 1) / 2, 2 * d);
}
int main() {
  long long q, i, j;
  scanf("%I64d%I64d%I64d", &n, &q, &m);
  for (p = 0, h = 1; h < n; p++, h *= 2)
    ;
  for (; q--;) {
    scanf("%I64d%I64d%I64d%I64d", &i, &j, &a, &b);
    i--;
    j--;
    a--;
    b--;
    if (a > n - 1 || i > n - 1)
      printf("0\n");
    else {
      if (b > n - 1) b = n - 1;
      if (j > n - 1) j = n - 1;
      i = rev(pos(i, n, 1));
      j = rev(pos(j, n, 1));
      printf("%I64d\n", gt(i, j, 0, h - 1));
    }
  }
  return 0;
}