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 read() {
  long long f = 1, x = 0;
  char c = getchar();
  while (c < '0' || '9' < c) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while ('0' <= c && c <= '9')
    x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  return f * x;
}
const int MAXN = 100000;
const int INF = 0x3f3f3f3f;
int Mod;
long long Cal(long long i, long long d, long long L, long long R, long long u) {
  if (i > u) return 0;
  long long n = R - L + 1, k = min(n - 1, max(0ll, (u - i) / d)),
            lst = i + k * d, ret = 0;
  if (!((i + lst) & 1))
    ret = (i + lst) / 2 % Mod * ((k + 1) % Mod) % Mod;
  else
    ret = (k + 1) / 2 % Mod * ((i + lst) % Mod) % Mod;
  return ret;
}
long long Query(long long i, long long d, long long L, long long R,
                long long qL, long long qR, long long u, long long v) {
  if (qL <= L && R <= qR)
    return (Cal(i, d, L, R, v) - Cal(i, d, L, R, u - 1) + Mod) % Mod;
  long long Mid = (L + R) >> 1, ret = 0;
  if (qL <= Mid) ret = (ret + Query(i, 2 * d, L, Mid, qL, qR, u, v)) % Mod;
  if (Mid + 1 <= qR)
    ret = (ret + Query(i + d, 2 * d, Mid + 1, R, qL, qR, u, v)) % Mod;
  return ret;
}
int main() {
  long long n = read();
  int m = read();
  Mod = read();
  for (int i = 1; i <= m; i++) {
    long long l = read(), r = read(), u = read(), v = read();
    printf("%lld\n", Query(1, 1, 1, n, l, r, u, v));
  }
  return 0;
}