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 INF = 1e9;
const double EPS = 1e-9;
int mod;
long long sum(long long n, long long r, long long v, long long &cnt) {
  if (r < 1) return 0;
  if (n == r) {
    v %= mod;
    long long rez = ((1 + v) * v / 2) % mod;
    cnt = v;
    return rez % mod;
  }
  long long m = (n + 1) / 2;
  if (r <= m) {
    long long x;
    long long rez = sum(m, r, (v + 1) / 2, x);
    cnt = x;
    return (rez * 2 - x) % mod;
  }
  long long x1;
  long long r1 = sum(m, m, (v + 1) / 2, x1);
  r1 = (r1 * 2 - x1) % mod;
  long long x2;
  long long r2 = sum(n - m, r - m, v / 2, x2);
  cnt = x1 + x2;
  return (r2 * 2 + r1) % mod;
}
long long calc(long long n, long long l, long long r, long long v) {
  l = min(l, n);
  r = min(r, n);
  v = min(v, n);
  if (v < 1) return 0;
  long long a, b;
  long long rez = sum(n, r, v, a) - sum(n, l - 1, v, b);
  while (rez < 0) rez += mod;
  return rez;
}
int main() {
  long long n;
  int m;
  scanf("%I64d%d%d", &n, &m, &mod);
  for (int i = 0; i < m; ++i) {
    long long l, r, u, v;
    scanf("%I64d%I64d%I64d%I64d", &l, &r, &u, &v);
    long long rez = (calc(n, l, r, v) - calc(n, l, r, u - 1)) % mod;
    if (rez < 0) rez += mod;
    printf("%I64d\n", rez);
  }
  return 0;
}