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 m, MOD;
long long a, b;
long long add(long long x, long long y) { return (x + y) % MOD; }
long long mult(long long x, long long y) { return (x * y) % MOD; }
long long query(long long mod, long long residue, long long block_start,
                long long block_end, long long l, long long r) {
  if (l == block_start && r == block_end) {
    long long lower_index;
    if (a < residue)
      lower_index = 0;
    else
      lower_index = (a - residue + mod - 1) / mod;
    if (b < residue) return 0;
    long long upper_index = (b - residue) / mod;
    if (lower_index > upper_index) return 0;
    mod %= MOD;
    residue %= MOD;
    long long i = lower_index % MOD;
    long long j = upper_index % MOD;
    long long x = add(mult(i, mod), mult(j, mod));
    x = add(x, residue);
    x = add(x, residue);
    x = x * (j - i + 1) / 2;
    x %= MOD;
    while (x < 0) x += MOD;
    return x;
  } else {
    long long mid = (block_start + block_end) / 2;
    if (r <= mid)
      return query(2 * mod, residue, block_start, mid, l, r);
    else if (l > mid)
      return query(2 * mod, residue + mod, mid + 1, block_end, l, r);
    else
      return (query(2 * mod, residue, block_start, mid, l, mid) +
              query(2 * mod, residue + mod, mid + 1, block_end, mid + 1, r)) %
             MOD;
  }
}
int main() {
  long long n;
  scanf("%I64d %I64d %I64d", &n, &m, &MOD);
  MOD *= 2;
  for (int i = 0; i < m; i++) {
    long long l, r;
    scanf("%I64d %I64d %I64d %I64d", &l, &r, &a, &b);
    if (a > n)
      printf("0\n");
    else {
      if (b > n) b = n;
      printf("%I64d\n", query(1, 1, 1, n, l, r) % (MOD / 2));
    }
  }
}