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 u, v;
int modVal, m;
long long n;
inline long long ceilDiv(long long a, long long b) {
  if (a % b == 0)
    return a / b;
  else if ((a >= 0 && b >= 0) || (a <= 0 && b <= 0))
    return a / b + 1;
  else
    return a / b;
}
inline long long floorDiv(long long a, long long b) {
  if (a % b == 0 || (a >= 0 && b >= 0) || (a <= 0 && b <= 0))
    return a / b;
  else
    return a / b - 1;
}
long long ask(long long a1, long long d, long long n, long long l,
              long long r) {
  if (l == 1 && r == n) {
    long long p = 1 + ceilDiv(u - a1, d), q = 1 + floorDiv(v - a1, d);
    p = max(1LL, p);
    q = min(n, q);
    if (p > q) return 0LL;
    long long ap = a1 + (p - 1) * d, cnt = q - p + 1, val;
    if (cnt & 1)
      val = ((((((ap) + modVal) % modVal) +
               (((((((q - p) / 2) + modVal) % modVal) *
                  (((d) + modVal) % modVal)) +
                 modVal) %
                modVal)) +
              modVal) %
             modVal);
    else
      cnt >>= 1,
          val = ((((((ap) + modVal) % modVal) * 2 +
                   (((q - p) + modVal) % modVal) * (((d) + modVal) % modVal)) +
                  modVal) %
                 modVal);
    return (
        (((((cnt) + modVal) % modVal) * (((val) + modVal) % modVal)) + modVal) %
        modVal);
  }
  long long mid = ceilDiv(n, 2);
  long long res = 0;
  if (l <= mid) res += ask(a1, d << 1, mid, l, min(r, mid));
  if (r > mid) res += ask(a1 + d, d << 1, n - mid, max(l - mid, 1LL), r - mid);
  return (((res) + modVal) % modVal);
}
int main() {
  scanf(
      "%lld"
      "%d%d",
      &n, &m, &modVal);
  while (m--) {
    long long l, r;
    scanf(
        "%lld"
        "%lld"
        "%lld"
        "%lld",
        &l, &r, &u, &v);
    printf(
        "%lld"
        "\n",
        ask(1, 1, n, l, r));
  }
}