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;
int modulo;
int getAns(long long start, long long delta, long long n, long long left,
           long long right, long long lower, long long upper) {
  if (n == 1) {
    return (lower <= start && start <= upper) ? start % modulo : 0;
  } else if (left == 0 && right == n - 1) {
    long long end = start + (n - 1) * delta;
    if (max(start, lower) <= min(end, upper)) {
      long long from;
      if (lower <= start) {
        from = start;
      } else {
        long long mod = (lower - start) % delta;
        from = (mod == 0 ? lower : lower + (delta - mod));
      }
      long long to;
      if (end <= upper) {
        to = end;
      } else {
        to = upper - (upper - start) % delta;
      }
      if (from <= to) {
        long long cnt = (to - from) / delta + 1;
        if (cnt % 2 == 0) {
          return (from + to) % modulo * (cnt / 2 % modulo) % modulo;
        } else {
          return (from + to) / 2 % modulo * (cnt % modulo) % modulo;
        }
      }
    }
    return 0;
  } else {
    long long mid = n / 2 + n % 2;
    long long ans = 0;
    if (left < mid) {
      ans = getAns(start, delta * 2, mid, left, min(right, mid - 1), lower,
                   upper);
    }
    if (mid <= right) {
      ans += getAns(start + delta, delta * 2, n - mid, max(left, mid) - mid,
                    right - mid, lower, upper);
      if (ans >= modulo) {
        ans -= modulo;
      }
    }
    return (int)ans;
  }
}
int main() {
  long long n;
  int m;
  while (scanf("%I64d%d%d", &n, &m, &modulo) != EOF) {
    while (m-- > 0) {
      long long left, right, lower, upper;
      scanf("%I64d%I64d%I64d%I64d", &left, &right, &lower, &upper);
      left--;
      right--;
      printf("%d\n", getAns(1, 1, n, left, right, lower, upper));
    }
  }
  return 0;
}