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, m, mod, sum;
void q(long long l, long long r, long long u, long long v, long long mm,
       long long mr, long long id, long long num) {
  long long mid = id + (num - 1) / 2;
  if (l == id && r == id + num - 1) {
    long long st, ed;
    if (u % mm > mr % mm)
      st = (u / mm + 1) * mm + mr % mm;
    else
      st = (u / mm) * mm + mr % mm;
    if (v % mm < mr % mm)
      ed = (v / mm - 1) * mm + mr % mm;
    else
      ed = (v / mm) * mm + mr % mm;
    long long ed1;
    if (n % mm < mr % mm)
      ed1 = (n / mm - 1) * mm + mr % mm;
    else
      ed1 = (n / mm) * mm + mr % mm;
    ed = ed < ed1 ? ed : ed1;
    if (st > ed) return;
    long long d = (ed - st) / mm + 1, a = ed + st;
    if (d % 2 == 0)
      d /= 2;
    else
      a /= 2;
    d %= mod;
    a %= mod;
    sum += (a * d) % mod;
    sum %= mod;
  } else if (l > mid)
    q(l, r, u, v, mm * 2, mr + mm, mid + 1, num - (num + 1) / 2);
  else if (r <= mid)
    q(l, r, u, v, mm * 2, mr, id, (num + 1) / 2);
  else {
    q(l, mid, u, v, mm * 2, mr, id, (num + 1) / 2);
    q(mid + 1, r, u, v, mm * 2, mr + mm, id + (num + 1) / 2,
      num - (num + 1) / 2);
  }
}
int main() {
  while (scanf("%I64d%I64d%I64d", &n, &m, &mod) != EOF) {
    while (m--) {
      long long l, r, u, v;
      scanf("%I64d%I64d%I64d%I64d", &l, &r, &u, &v);
      sum = 0;
      q(l - 1, r - 1, u, v, 1, 1, 0, n);
      printf("%I64d\n", sum);
    }
  }
  return 0;
}