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>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
template <class T>
T abs(const T& a) {
  return a > 0 ? a : -a;
}
template <class T>
int sgn(const T& a) {
  return a > 0 ? 1 : (a < 0 ? -1 : 0);
}
const double PI = 2.0 * acos(1.0);
long long n, mod;
long long oddn(long long a) { return a % 2 ? a : a + 1; }
long long oddp(long long a) { return a % 2 ? a : a - 1; }
long long evnn(long long a) { return a % 2 ? a + 1 : a; }
long long evnp(long long a) { return a % 2 ? a - 1 : a; }
long long get(long long fst, long long lst, long long k, long long l,
              long long r, long long L, long long R, long long shift) {
  if (l > r) return 0;
  if (shift + 1 > R || shift + k < L) return 0;
  long long d = (k > 1 ? (lst - fst) / (k - 1) : 1);
  if (L <= 1 + shift && k + shift <= R) {
    long long A = 2 * fst + d * (l - 1) + d * (r - 1);
    long long B = (r - l + 1);
    if (A % 2 == 0)
      A /= 2;
    else
      B /= 2;
    A %= mod;
    B %= mod;
    return (A * B) % mod;
  }
  long long s1 = get(fst, k % 2 ? lst : lst - d, (k + 1) / 2, (oddn(l) + 1) / 2,
                     (oddp(r) + 1) / 2, L, R, shift);
  long long s2 = get(fst + d, k % 2 ? lst - d : lst, k / 2, evnn(l) / 2,
                     evnp(r) / 2, L, R, shift + (k + 1) / 2);
  return (s1 + s2) % mod;
}
int main() {
  int m;
  cin >> n >> m >> mod;
  for (int i = 0; i < m; i++) {
    long long l, r, u, v;
    cin >> l >> r >> u >> v;
    if (u > n)
      cout << 0 << endl;
    else {
      v = min(v, n);
      cout << get(1, n, n, u, v, l, r, 0) << endl;
    }
  }
  return 0;
}