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;
const double pi = 3.1415926535;
const double eps = 1e-6;
long long n, m, mod, l, r, u, v;
long long work(long long l, long long r, long long p, long long q, long long u,
               long long v, long long s, long long d) {
  if (p <= l && r <= q) {
    long long t = min(s + (r - l - 1) * d, s + (v - s) / d * d);
    if (s + (r - l - 1) * d < u) return 0;
    s = s + (u - s) / d * d;
    while (s < u) s += d;
    if (s > v || t < u) return 0;
    long long N = (t - s) / d + 1;
    long long A = s + t;
    if (N % 2 == 0)
      N /= 2;
    else
      A /= 2;
    N %= mod, A %= mod;
    return (N * A) % mod;
  } else {
    long long mi = (l + r + 1) / 2;
    long long S = 0;
    if (p < mi) S += work(l, mi, p, q, u, v, s, d << 1);
    if (S >= mod) S -= mod;
    if (mi < q) S += work(mi, r, p, q, u, v, s + d, d << 1);
    if (S >= mod) S -= mod;
    return S;
  }
}
int main() {
  cin >> n >> m >> mod;
  for (int i = 0; i < m; i++) {
    cin >> l >> r >> u >> v;
    cout << work(0, n, l - 1, r, u, v, 1, 1) << endl;
  }
}