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 n;
long long m;
long long u, v, x, y;
long long MOD(long long _x) { return (_x + m) % m; }
long long sum(long long a, long long b, long long c) {
  if (c < b) return 0;
  c -= b;
  long long re = MOD(MOD(c / a + 1) * MOD(b));
  long long ct = c / a;
  MOD(ct);
  if (ct % 2 == 0)
    re += MOD(MOD(a) * MOD(MOD(ct / 2) * MOD(ct + 1)));
  else
    re += MOD(MOD(a) * MOD(MOD(ct) * MOD((ct + 1) / 2)));
  return re;
}
long long f1(long long f, long long l, long long pre, long long st) {
  if (l < u or f > v)
    return 0;
  else if (u <= f and l <= v)
    return MOD(sum(pre, st, y) - sum(pre, st, x - 1));
  return MOD(f1(f, (f + l) / 2, 2 * pre, st) +
             f1((f + l) / 2 + 1, l, 2 * pre, pre + st));
}
void init() {
  int q;
  scanf("%I64d %d %I64d", &n, &q, &m);
  for (int i = 0; i < q; i++) {
    scanf("%I64d %I64d %I64d %I64d", &u, &v, &x, &y);
    if (y > n) y = n;
    if (x > n)
      printf("0\n");
    else
      printf("%I64d\n", MOD(f1(1, n, 1, 1)));
  }
}
int main() { init(); }