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>
long long l, r, u, v;
long long N, ret, MOD, M, i, j, k, one = 1;
void count(long long left, long long right, long long ql, long long qr,
           long long lv, long long D) {
  if (left > right || ql > qr) return;
  long long n, m, rv;
  if (left == ql && right == qr) {
    n = right - left + 1;
    rv = lv + ((n - 1) << D);
    ql = (((u - lv + (one << D) - 1) >> D) << D) + lv;
    qr = (((v - lv) >> D) << D) + lv;
    ql = (ql <= lv ? lv : ql);
    qr = (qr >= rv ? rv : qr);
    if (!(u <= ql && ql <= qr && qr <= v)) return;
    n = ((qr - ql) >> D) + 1;
    ret += (((ql + qr) % MOD) * (((n - (n & 1)) >> 1) % MOD));
    ret += (((ql + qr) >> 1) * (n & 1));
    ret %= MOD;
    return;
  }
  m = (left + right) >> 1;
  n = (m <= qr ? m : qr);
  if (ql <= m) count(left, m, ql, n, lv, D + 1);
  n = (m + 1 >= ql ? m + 1 : ql);
  if (m < qr) count(m + 1, right, n, qr, lv + (one << D), D + 1);
  return;
}
int main() {
  scanf("%I64d %I64d %I64d", &N, &M, &MOD);
  for (i = 0; i < M; i++) {
    scanf("%I64d %I64d %I64d %I64d", &l, &r, &u, &v);
    ret = 0;
    count(1, N, l, r, 1, 0);
    printf("%I64d\n", ret);
  }
  return 0;
}