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, l, r, u, v, ans;
int MOD, m;
void updata(long long num, long long d, long long cnt) {
  if (cnt == 1) {
    if (num >= u && num <= v) ans = (ans + num % MOD) % MOD;
    return;
  }
  long long l = (u - num) / d, r = (v - num) / d;
  if (num + l * d < u) l++;
  if (num + r * d > v) r--;
  l = max(0ll, l);
  r = min(r, cnt - 1);
  if (l > r) return;
  long long g = (num + l * d + num + r * d), h = (r - l + 1);
  if (g % 2 == 0)
    g /= 2;
  else
    h /= 2;
  g %= MOD;
  h %= MOD;
  ans += g * h % MOD;
  ans %= MOD;
}
void solve(long long st, long long ed, long long num, long long d, long long l,
           long long r) {
  if (l == st && r == ed) {
    updata(num, d, ed - st + 1);
    return;
  }
  long long mid = (st + ed) / 2;
  if (r <= mid)
    solve(st, mid, num, d * 2, l, r);
  else if (l > mid)
    solve(mid + 1, ed, num + d, d * 2, l, r);
  else {
    solve(st, mid, num, d * 2, l, mid);
    solve(mid + 1, ed, num + d, d * 2, mid + 1, r);
  }
}
int main() {
  scanf("%I64d%d%d", &n, &m, &MOD);
  while (m--) {
    scanf("%I64d%I64d%I64d%I64d", &l, &r, &u, &v);
    ans = 0;
    solve(1ll, n, 1ll, 1ll, l, r);
    printf("%I64d\n", ans);
  }
}