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;
int m, mod;
long long solve(long long ql, long long qr, long long u, long long xl = 0,
                long long xr = n, long long val = 0, int t = 0) {
  if (xr <= ql || qr <= xl) return 0;
  if (ql <= xl && xr <= qr) {
    u--;
    if (u - val < 0) return 0;
    long long k = 1 + (u - val) / (1LL << t);
    long long kp = 2 * val + ((k - 1) * (1LL << t));
    long long ret = k;
    if (kp & 1)
      k >>= 1;
    else
      kp >>= 1;
    ret += k % mod * (kp % mod) % mod;
    if (ret >= mod) ret -= mod;
    return ret;
  }
  long long xm = (xl + xr + 1) >> 1;
  long long ret = solve(ql, qr, u, xl, xm, val, t + 1) +
                  solve(ql, qr, u, xm, xr, val + (1LL << t), t + 1);
  if (ret >= mod) ret -= mod;
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> mod;
  for (int i = 0; i < m; i++) {
    long long l, r, u, v;
    cin >> l >> r >> u >> v;
    l--;
    long long ret = solve(l, r, min(n, v)) - solve(l, r, min(n, u - 1));
    ret = (ret + mod) % mod;
    cout << ret << '\n';
  }
}