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 x = 0, F = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') F = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * F;
}
int MOD;
int add(int a, int b) { return (a + b >= MOD) ? a + b - MOD : a + b; }
int dec(int a, int b) { return (a - b < 0) ? a - b + MOD : a - b; }
int mul(long long a, long long b) {
  return (1LL * (a % MOD) * (b % MOD)) % MOD;
}
int calc(long long a, long long d, long long num, long long lim) {
  num = min(num, (lim + d - a) / d);
  long long sum;
  if (num & 1)
    sum = mul(num, (num - 1) / 2);
  else
    sum = mul(num / 2, num - 1);
  return add(mul(a, num), mul(d, sum));
}
int Query(long long l, long long r, long long L, long long R, long long u,
          long long v, long long a, long long d) {
  if (r < L || R < l) return 0;
  if (L <= l && r <= R)
    return dec(calc(a, d, r - l + 1, v), calc(a, d, r - l + 1, u - 1));
  return add(Query(l, (l + r >> 1), L, R, u, v, a, d << 1),
             Query((l + r >> 1) + 1, r, L, R, u, v, a + d, d << 1));
}
int main() {
  long long n = read();
  int q = read();
  MOD = read();
  while (q--) {
    long long L = read(), R = read(), u = read(), v = read();
    printf("%d\n", Query(1, n, L, R, u, v, 1, 1));
  }
}