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;
int cond = 1;
int MD;
long long countS(long long suf, long long p2, long long n) {
  if (n <= suf) {
    return 0LL;
  }
  return (n - suf + p2 - 1) / p2;
}
long long mn(long long a, long long b) { return (a % MD) * (b % MD) % MD; }
long long sum(long long c) {
  if (c % 2) {
    return mn(c, (c - 1) / 2);
  } else {
    return mn(c / 2, c - 1);
  }
}
long long F(long long suf, long long p2, long long cnt, long long n) {
  if (cnt == 0) {
    return 0;
  }
  if (countS(suf, p2, n) == cnt) {
    return (mn(sum(cnt), p2) + mn(suf, cnt) + cnt) % MD;
  } else {
    if (countS(suf, p2 * 2, n) >= cnt) {
      return F(suf, p2 * 2, cnt, n);
    } else {
      return (F(suf, p2 * 2, countS(suf, p2 * 2, n), n) +
              F(suf + p2, p2 * 2, cnt - countS(suf, p2 * 2, n), n)) %
             MD;
    }
  }
}
long long findUB(long long cnt, long long m, long long n) {
  long long p2 = 1;
  long long suf = 0;
  long long rem = cnt;
  long long rcnt = 0;
  for (int i = 0; i < 61; ++i) {
    if (countS(suf, p2 * 2, n) < rem) {
      rcnt += countS(suf, p2 * 2, m);
      rem -= countS(suf, p2 * 2, n);
      suf += p2;
    }
    p2 *= 2;
  }
  rcnt += countS(suf, p2, m);
  return rcnt;
}
long long get(long long cnt, long long m, long long n) {
  if (cnt == 0 || m == 0) {
    return 0LL;
  }
  cnt = findUB(cnt, m, n);
  return F(0, 1, cnt, m);
}
long long n;
int main() {
  ios_base::sync_with_stdio(false);
  int m;
  cin >> n >> m >> MD;
  for (int i = 1; i <= m; ++i) {
    long long l, r, u, v;
    cin >> l >> r >> u >> v;
    if (u > n) {
      cout << "0\n";
      continue;
    }
    v = min(v, n);
    long long res = get(r, v, n) - get(r, u - 1, n) - get(l - 1, v, n) +
                    get(l - 1, u - 1, n);
    cout << ((res % MD) + MD) % MD << "\n";
  }
  return 0;
}