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>
#pragma comment(linker, "/STACK:134217728")
using namespace std;
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
int m;
long long n;
long long mod;
pair<long long, long long> getans(long long n, long long lp, long long rp,
                                  long long l, long long r) {
  l = max(l, 0ll);
  r = min(r, n - 1);
  lp = max(lp, 0ll);
  rp = min(rp, n);
  if (l >= n || r < 0 || l > r) return make_pair(0ll, 0ll);
  if (lp >= n || rp <= 0 || lp >= rp) return make_pair(0ll, 0ll);
  if (lp <= 0 && rp >= n) {
    long long a = (r - l + 1);
    long long b = (r + l);
    if (a % 2 == 0)
      a /= 2;
    else
      b /= 2;
    a %= mod;
    b %= mod;
    return make_pair((a * b) % mod, r - l + 1);
  }
  pair<long long, long long> a =
      getans((n + 1) / 2, lp, rp, (l + 1) / 2, r / 2);
  a.first *= 2;
  a.first %= mod;
  pair<long long, long long> b =
      getans(n / 2, lp - (n + 1) / 2, rp - (n + 1) / 2, l / 2, (r + 1) / 2 - 1);
  b.first = ((b.first * 2) + b.second) % mod;
  return make_pair((a.first + b.first) % mod, a.second + b.second);
}
int main() {
  cin >> n >> m >> mod;
  for (int i = 0; i < (m); i++) {
    long long lp, rp, l, r;
    cin >> lp >> rp >> l >> r;
    lp--;
    l--;
    r--;
    pair<long long, long long> res = getans(n, lp, rp, l, r);
    res.first += res.second;
    cout << res.first % mod << endl;
  }
  return 0;
}