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;
const double pi = acos(-1);
long double eps = 1e-9;
long long pp = 1000000007;
long long mo(long long a, long long pp) {
  if (a >= 0 && a < pp) return a;
  a %= pp;
  if (a < 0) a += pp;
  return a;
}
long long powmod(long long a, long long b, long long pp) {
  long long ans = 1;
  for (; b; b >>= 1, a = mo(a * a, pp))
    if (b & 1) ans = mo(ans * a, pp);
  return ans;
}
int dx[5] = {0, -1, 1, 0, 0}, dy[5] = {0, 0, 0, -1, 1};
long long read() {
  long long ans = 0;
  char last = ' ', ch = getchar();
  while (ch < '0' || ch > '9') last = ch, ch = getchar();
  while (ch >= '0' && ch <= '9') ans = ans * 10 + ch - '0', ch = getchar();
  if (last == '-') ans = -ans;
  return ans;
}
long long p;
long long calc(long long x, long long y, long long u, long long v, long long a,
               long long b) {
  bool flag = 0;
  long long num = (y - x + 1), ka = ((v - b) / a) * a + b,
            ki = ((u - b) / a) * a + b, t1, t2;
  if (v - b < 0LL) ka = ((v - b) / a - 1) * a + b;
  if (u - b < 0LL) ki = ((u - b) / a - 1) * a + b;
  ka = min(ka, num * a - a + b);
  ki = max(b, ki);
  if (ki < u) ki += a;
  if (ki > ka) return 0LL;
  if ((ka + ki) % 2LL == 0)
    t1 = (ka + ki) / 2LL, flag = 1;
  else
    t1 = (ka + ki);
  if (!flag && ((ka - ki) / a + 1) % 2LL == 0)
    t2 = ((ka - ki) / a + 1) / 2LL;
  else
    t2 = ((ka - ki) / a + 1);
  return ((t1 % p) * (t2 % p) % p);
}
long long solve(long long x, long long y, long long l, long long r, long long u,
                long long v, long long a, long long b) {
  if (l > r) return 0;
  if (l <= x && y <= r) {
    return calc(x, y, u, v, a, b) % p;
  }
  long long mid = (x + y) >> 1;
  if (r <= mid) return solve(x, mid, l, r, u, v, a + a, b) % p;
  if (l >= mid + 1) return solve(mid + 1, y, l, r, u, v, a + a, a + b) % p;
  return (solve(x, mid, l, r, u, v, a + a, b) +
          solve(mid + 1, y, l, r, u, v, a + a, a + b)) %
         p;
}
int main() {
  long long n = read(), m = read();
  p = read();
  while (m--) {
    long long l = read(), r = read(), u = read(), v = read();
    cout << solve(1, n, l, r, u, v, 1, 1) << endl;
  }
  return 0;
}