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;
long long u, v;
int m;
int mod;
long long calc(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long solve(long long node, long long left, long long right,
                long long nleft, long long nright, long long start,
                long long step) {
  if (right < nleft || left > nright) return 0;
  if (left >= nleft && right <= nright) {
    long long cur = left;
    long long x = start;
    if (x < u) {
      cur += (u - x + step - 1LL) / step;
      if (cur > right) return 0;
      x += (u - x + step - 1LL) / step * step;
    }
    if (x > v) return 0;
    long long cur2 = cur + (v - x) / step;
    cur2 = min(cur2, right);
    long long tmp = cur2 - cur;
    long long ret = calc(x, tmp + 1LL);
    if (tmp % 2LL == 0)
      ret += calc(calc(tmp / 2LL, tmp + 1LL), step);
    else
      ret += calc(calc(tmp, (tmp + 1LL) / 2LL), step);
    if (ret >= mod) ret -= mod;
    return ret;
  }
  long long mid = (left + right) / 2LL;
  long long res = 0;
  res += solve(node * 2LL + 1LL, left, mid, nleft, nright, start, step * 2LL);
  res += solve(node * 2LL + 2LL, mid + 1LL, right, nleft, nright, start + step,
               step * 2LL);
  if (res >= mod) res -= mod;
  return res;
}
int main() {
  cin >> n >> m >> mod;
  for (int i = 0; i < m; ++i) {
    long long l, r;
    cin >> l >> r >> u >> v;
    --l;
    --r;
    cout << solve(0, 0, n - 1, l, r, 1, 1) << endl;
  }
  return 0;
}