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 m;
pair<long long, long long> solve(long long n, long long r, long long u,
                                 long long v) {
  u = max(u, 1ll);
  v = min(v, n);
  if (r <= 0ll) return make_pair(0, 0);
  if (v < u) return make_pair(0, 0);
  if (n == 1ll) {
    if (u <= 1ll && v >= 1ll) return make_pair((1 % m), 1);
    return make_pair(0, 0);
  }
  pair<long long, long long> res = make_pair(0ll, 0ll), tmpp;
  long long tmp;
  if (r > ((n + 1ll) / 2ll)) {
    tmp = v;
    res.second = (v + 1ll) / 2ll;
    if (!(v & 1ll)) tmp--;
    tmp = (tmp / 2ll) + 1ll;
    tmp %= m;
    res.first = tmp * tmp;
    tmp = u - 1ll;
    res.second -= u / 2ll;
    if (tmp != 0ll) {
      if (!(tmp & 1ll)) tmp--;
      tmp = (tmp / 2ll) + 1ll;
      tmp %= m;
      res.first -= tmp * tmp;
    }
    tmpp = solve(n / 2ll, r - ((n + 1ll) / 2ll), (u + 1ll) / 2ll, v / 2ll);
    res.second += tmpp.second, res.first += 2ll * tmpp.first;
    res.first = ((res.first % m) + m) % m;
    return res;
  }
  res = solve((n + 1ll) / 2ll, r, (u + 2ll) / 2ll, (v + 1ll) / 2ll);
  res.first = (((res.first * 2ll - res.second) % m) + m) % m;
  return res;
}
int main() {
  long long n, l, r, u, v;
  int t;
  scanf("%I64d%d%I64d", &n, &t, &m);
  while (t > 0) {
    scanf("%I64d%I64d%I64d%I64d", &l, &r, &u, &v);
    printf(
        "%I64d\n",
        ((solve(n, r, u, v)).first - (solve(n, l - 1ll, u, v)).first + m) % m);
    t--;
  }
  return 0;
}