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 n, m, p;
long long getmax(long long a, long long b) { return a > b ? a : b; }
long long getmin(long long a, long long b) { return a < b ? a : b; }
long long getans(long long s, long long d, long long lower, long long upper,
                 long long st, long long ed, long long down, long long up) {
  if (lower == st && upper == ed) {
    down = getmax(s, down);
    up = getmin(s + (upper - lower) * d, up);
    if (up < down) return 0;
    long long p1, p2, t1 = (up - s + 1) / d + ((up - s + 1) % d != 0),
                      t2 = (down - s) / d + ((down - s) % d != 0);
    if (t1 & 1)
      p1 = ((t1 % p) * (((t1 - 1) / 2) % p)) % p;
    else
      p1 = (((t1 / 2) % p) * ((t1 - 1) % p)) % p;
    if (t2 & 1)
      p2 = ((t2 % p) * (((t2 - 1) / 2) % p)) % p;
    else
      p2 = (((t2 / 2) % p) * ((t2 - 1) % p)) % p;
    return ((t1 * (s % p) % p + p1 * (d % p) % p - t2 * (s % p) % p -
             p2 * (d % p) % p) %
                p +
            p) %
           p;
  }
  long long mid = (lower + upper) >> 1;
  if (ed <= mid)
    return getans(s, d * 2, lower, mid, st, ed, down, up);
  else if (st > mid)
    return getans(s + d, d * 2, mid + 1, upper, st, ed, down, up);
  else
    return (getans(s, d * 2, lower, mid, st, mid, down, up) +
            getans(s + d, d * 2, mid + 1, upper, mid + 1, ed, down, up)) %
           p;
}
int main() {
  cin >> n >> m >> p;
  while (m--) {
    long long l, r, u, v;
    cin >> l >> r >> u >> v;
    cout << getans(1, 1, 1, n, l, r, u, v) << endl;
  }
  return 0;
}