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;
template <class T>
inline void read(T& x) {
  bool fu = 0;
  char c;
  for (c = getchar(); c <= 32; c = getchar())
    ;
  if (c == '-') fu = 1, c = getchar();
  for (x = 0; c > 32; c = getchar()) x = x * 10 + c - '0';
  if (fu) x = -x;
};
template <class T>
inline void read(T& x, T& y) {
  read(x);
  read(y);
}
template <class T>
inline void read(T& x, T& y, T& z) {
  read(x);
  read(y);
  read(z);
}
inline char getc() {
  char c;
  for (c = getchar(); c <= 32; c = getchar())
    ;
  return c;
}
long long n, m, i, j, k, l, r, u, v, p, mod;
long long ans, inv2;
long long pow(long long a, long long b, long long c) {
  long long ans = 1;
  for (; b; b >>= 1, a = a * a % c)
    if (b & 1) ans = ans * a % c;
  return ans;
}
long long calc(long long st, long long d, long long t, long long y) {
  if (y < st) return 0;
  long long z = min(t, (y + d - st) / d);
  long long tt1 = st + (z - 1) * d + st, tt2 = z;
  if (tt1 % 2 == 0)
    tt1 /= 2;
  else
    tt2 /= 2;
  return (tt1 % mod) * (tt2 % mod) % mod;
}
void go(long long len, long long st, long long d, long long x, long long y) {
  if (x == 0) return;
  long long t = (len + 1) / 2;
  if (x >= t) {
    ans = ((ans) + (calc(st, d * 2, t, y))) % mod;
    go(len - t, st + d, d * 2, x - t, y);
  } else {
    go(t, st, d * 2, x, y);
  }
}
long long c(long long x, long long y) {
  ans = 0;
  go(n, 1, 1, x, y);
  return ans;
}
int main() {
  read(n, m, mod);
  for (; m; m--) {
    read(l, r);
    read(u, v);
    long long res;
    res = c(r, v) + c(l - 1, u - 1) - c(r, u - 1) - c(l - 1, v);
    res = (res % mod + mod) % mod;
    printf("%I64d\n", res);
  }
  return 0;
}