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;
unsigned long long n, m, MOD;
unsigned long long mn, mx;
unsigned long long ans;
inline unsigned long long qb(unsigned long long a, unsigned long long b,
                             unsigned long long MOD) {
  unsigned long long ret = 0;
  while (b) {
    if (b & 1) ret = (ret + a) % MOD;
    a += a;
    a %= MOD;
    b >>= 1;
  }
  return ret;
}
inline void getans(unsigned long long he, unsigned long long N,
                   unsigned long long d, unsigned long long r) {
  unsigned long long st, ed;
  if (he >= mn)
    st = 1;
  else
    st = (mn - he - 1) / d + 2;
  if (mx >= he + (N - 1) * d)
    ed = N;
  else
    ed = (mx - he) / d + 1;
  unsigned long long a = he * 2 + (st + ed - 2) * d;
  unsigned long long b = ed - st + 1;
  if (a & 1)
    b >>= 1;
  else
    a >>= 1;
  ans = (ans + ((a % MOD) * (b % MOD)) % MOD) % MOD;
}
inline void calc(unsigned long long he, unsigned long long N,
                 unsigned long long d, unsigned long long l,
                 unsigned long long r) {
  if (he > mx) return;
  if (he + (N - 1) * d < mn) return;
  if (l == 1 && r == N) {
    getans(he, N, d, r);
    return;
  }
  unsigned long long mid = (N + 1) >> 1;
  if (r <= mid)
    calc(he, mid, d * 2, l, r);
  else if (l > mid)
    calc(he + d, N - mid, d * 2, l - mid, r - mid);
  else
    calc(he, mid, d * 2, l, mid), calc(he + d, N - mid, d * 2, 1, r - mid);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m >> MOD;
  while (m--) {
    unsigned long long l, r;
    cin >> l >> r >> mn >> mx;
    ans = 0;
    calc(1, n, 1, l, r);
    cout << ans << endl;
  }
  return 0;
}