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 int K = 3010;
const int N = 5010;
const int P = 100;
const int B = 11;
const int M = 1000000000;
long long n, m;
int md;
int calc(long long x, long long r, long long s, long long a) {
  a -= x;
  if (a < 0) return 0;
  long long ct = a / r + 1;
  ct = min(ct, s);
  long long sm = ct % 2 == 0 ? ((ct / 2) % md) * ((ct - 1) % md)
                             : (ct % md) * (((ct - 1) / 2) % md);
  sm %= md;
  long long ans = (((ct % md) * (x % md)) % md + (sm * (r % md)) % md) % md;
  return ans;
}
int solve(long long x, long long r, long long il, long long ir, long long ql,
          long long qr, long long mn, long long mx) {
  if (ql <= il && ir <= qr) {
    int ans = calc(x, r, ir - il + 1, mx) - calc(x, r, ir - il + 1, mn - 1);
    if (ans < 0) ans += md;
    ans %= md;
    return ans;
  }
  if (ir < ql || il > qr) return 0;
  long long rt = (ir - il + 1) / 2;
  long long lt = (ir - il + 1) - rt;
  int ans = solve(x, r * 2LL, il, il + lt - 1, ql, qr, mn, mx);
  if (rt > 0) ans += solve(x + r, r * 2LL, il + lt, ir, ql, qr, mn, mx);
  ans %= md;
  return ans;
}
long long l, r, x, y;
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> m >> md;
  for (int i = 1; i <= m; ++i) {
    cin >> l >> r >> x >> y;
    cout << solve(1, 1, 1, n, l, r, x, y) << '\n';
  }
}