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;
int m, mod;
long long q_l, q_r, u, v;
long long whole(long long start, long long end, long long upper) {
  if (upper < start) return 0;
  long long ct = (upper - start) / end + 1;
  upper = start * 2 + (ct - 1) * end;
  if (ct % 2 == 0)
    ct >>= 1;
  else
    upper >>= 1;
  return ct % mod * (upper % mod) % mod;
}
long long query(long long cur_l, long long cur_r, long long st_start,
                long long st_end) {
  if (cur_l >= q_l && cur_r <= q_r)
    return (whole(st_start, st_end, v) - whole(st_start, st_end, u - 1) + mod) %
           mod;
  long long mid = (cur_l + cur_r) >> 1, ans = 0;
  if (mid >= q_l) (ans += query(cur_l, mid, st_start, st_end * 2)) %= mod;
  if (mid + 1 <= q_r)
    (ans += query(mid + 1, cur_r, st_start + st_end, st_end * 2)) %= mod;
  return ans;
}
int main() {
  scanf("%I64d%d%d", &n, &m, &mod);
  do {
    scanf("%I64d%I64d%I64d%I64d", &q_l, &q_r, &u, &v);
    if (v > n) v = n;
    if (u > v)
      puts("0");
    else
      printf("%I64d\n", query(1, n, 1, 1));
  } while (--m);
  return 0;
}