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 n, m, mod;
long long sum(long long x) {
  if (x % 2 == 0)
    return (x / 2 % mod) * ((x - 1) % mod) % mod;
  else
    return ((x - 1) / 2 % mod) * (x % mod) % mod;
}
pair<long long, long long> get(long long n, long long p, long long x) {
  if (p == 0 || x < 0) return make_pair(0, 0);
  if (p < (n + 1) / 2) {
    pair<long long, long long> out = get((n + 1) / 2, p, x >> 1);
    out.second *= 2;
    out.second %= mod;
    return out;
  }
  long long cnt = min((n + 1) / 2, (x >> 1) + 1);
  long long ans = sum(cnt) % mod * 2 % mod;
  if (!(x & 1)) x--;
  pair<long long, long long> out = get(n / 2, p - (n + 1) / 2, x >> 1);
  out.second *= 2;
  out.second %= mod;
  out.second += out.first;
  out.second %= mod;
  out.first += cnt;
  out.first %= mod;
  out.second += ans;
  out.second %= mod;
  return out;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m >> mod;
  while (m--) {
    long long l, r, u, v;
    cin >> l >> r >> u >> v;
    u--;
    v--;
    l--;
    r--;
    pair<long long, long long> tmp;
    tmp = get(n, r + 1, v);
    long long ans = (tmp.second + tmp.first) % mod;
    tmp = get(n, r + 1, u - 1);
    ans -= (tmp.second + tmp.first) % mod;
    ans %= mod, ans += mod, ans %= mod;
    tmp = get(n, l, v);
    long long bad = (tmp.second + tmp.first) % mod;
    tmp = get(n, l, u - 1);
    bad -= (tmp.first + tmp.second) % mod;
    bad %= mod, bad += mod, bad %= mod;
    ans -= bad;
    ans %= mod, ans += mod, ans %= mod;
    cout << ans << "\n";
  }
  return 0;
}