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, mod, l, r, u, v;
long long go(long long b, long long le, long long numb, long long nume) {
  if (nume < l || numb > r) return 0;
  if (numb >= l && nume <= r) {
    long long k = (v - b) / le;
    k = min(k, (long long)(nume - numb));
    k %= mod;
    long long ans1 = 0;
    if (v >= b) {
      long long p1;
      if (k % 2)
        p1 = ((k % mod) * ((k + 1) / 2 % mod)) % mod;
      else
        p1 = ((k / 2 % mod) * ((k + 1) % mod)) % mod;
      ans1 =
          ((b % mod) * ((k + 1) % mod) % mod + (le % mod) * (p1 % mod)) % mod;
    }
    long long k1 = (u - 1 - b) / le;
    k1 = min(k1, (long long)(nume - numb));
    k1 %= mod;
    long long ans2 = 0;
    if (u - 1 >= b) {
      long long p2;
      if (k1 % 2)
        p2 = ((k1 % mod) * ((k1 + 1) / 2 % mod)) % mod;
      else
        p2 = ((k1 / 2 % mod) * ((k1 + 1) % mod)) % mod;
      ans2 =
          ((b % mod) * ((k1 + 1) % mod) % mod + (le % mod) * (p2 % mod)) % mod;
    }
    return (ans1 - ans2 + 2 * mod) % mod;
  } else {
    long long ans = 0;
    ans += go(b, 2 * le, numb, (numb + nume) / 2);
    ans += go(b + le, 2 * le, (numb + nume) / 2 + 1, nume);
    return ans % mod;
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m >> mod;
  for (long long i = 1; i <= m; i++) {
    cin >> l >> r >> u >> v;
    cout << go(1, 1, 1, n) << endl;
  }
  return 0;
}