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 mdc(long long a, long long b) { return (b ? mdc(b, a % b) : a); }
long long f(long long u, long long v, long long b, long long delta, long long n,
            long long wanted, long long mod) {
  long long n1, n2;
  if (!wanted) return 0;
  if (n != wanted) {
    n1 = n / 2 + (n % 2 ? 1 : 0), n2 = n / 2;
    if (n1 >= wanted) return f(u, v, b, delta * 2, n1, wanted, mod);
    return (f(u, v, b, delta * 2, n1, n1, mod) +
            f(u, v, b + delta, delta * 2, n2, wanted - n1, mod)) %
           mod;
  }
  n1 = (u + delta - b) / delta;
  if (mdc(u + delta - b, delta) != delta) ++n1;
  if ((u + delta - b) > delta * n) return 0;
  if ((u + delta - b) < delta) n1 = 1;
  n2 = (v + delta - b) / delta;
  if ((v + delta - b) < delta) return 0;
  if ((v + delta - b) > delta * n) n2 = n;
  n = n2 - n1 + 1;
  if (delta == 1) {
    if ((n2 - n1) % 2 == 0)
      return ((b + (((((n2 + n1 - 2) / 2) % mod) * delta) % mod)) * (n % mod)) %
             mod;
    return (((2 * b) % mod + ((delta % mod) * ((n2 + n1 - 2) % mod)) % mod) *
            ((n / 2) % mod)) %
           mod;
  }
  return ((((b % mod) + ((delta / 2) % mod) * ((n2 + n1 - 2) % mod)) % mod) *
          (n % mod)) %
         mod;
}
int main() {
  long long n, m, mod;
  cin >> n >> m >> mod;
  for (int i = 0; i < m; ++i) {
    long long l, r, u, v;
    cin >> l >> r >> u >> v;
    long long r1 = f(u, v, 1, 1, n, r, mod), r2 = f(u, v, 1, 1, n, l - 1, mod);
    cout << (r1 - r2 + mod) % mod << endl;
  }
  return 0;
}