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 MOD;
long long n;
long long add(long long a, long long b) { return (a + b + MOD) % MOD; }
long long mul(long long a, long long b) { return (a * b + MOD) % MOD; }
long long get_sum(long long len, long long a, long long d) {
  len = min(len, n);
  if (a > len) return 0;
  long long k = (len - a) / d;
  k %= MOD;
  d %= MOD;
  return (mul(k + 1, a) + mul(d, (k * (k + 1) / 2) % MOD) + MOD) % MOD;
}
long long calc(long long tl, long long tr, long long l, long long r,
               long long u, long long v, long long a, long long d) {
  l = max(l, tl);
  r = min(r, tr);
  if (l > r) return 0;
  if (tl == l && tr == r) {
    long long s1 = get_sum(v, a, d);
    long long s2 = get_sum(u - 1, a, d);
    return (s1 - s2 + MOD) % MOD;
  } else {
    long long a1 = calc(tl, tl + (tr - tl) / 2, l, r, u, v, a, d * 2);
    long long a2 = calc(tl + (tr - tl) / 2 + 1, tr, l, r, u, v, a + d, d * 2);
    return add(a1, a2);
  }
}
void solve() {
  long long m;
  cin >> n >> m >> MOD;
  long long l, r, u, v;
  for (int i = 0; i < m; i++) {
    scanf("%I64d%I64d%I64d%I64d", &l, &r, &u, &v);
    long long ans = calc(1, n, l, r, u, v, 1, 1);
    printf("%I64d\n", ans);
  }
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}