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;
const int INF = 2147483647;
const long long LLINF = 9223372036854775807LL;
int mod;
int getsum(long long x) {
  if (x == 0) return 0;
  x %= mod;
  return (x * (x + 1) / 2) % mod;
}
pair<int, int> query(long long n, long long l, long long r, long long u,
                     long long v, long long L, long long R) {
  if (l > R || L > r) return make_pair(0, 0);
  if (L >= l && R <= r) {
    u = max(u, 1LL);
    v = min(v, n);
    if (u > v) return make_pair(0, 0);
    int ans = (getsum(v) - getsum(u - 1) + mod) % mod;
    return make_pair(ans, (int)((v - u + 1) % mod));
  }
  long long m = (L + R) >> 1;
  pair<int, int> q1 = query((n + 1) / 2, l, r, u / 2 + 1, (v + 1) / 2, L, m);
  pair<int, int> q2 = query(n / 2, l, r, (u + 1) / 2, v / 2, m + 1, R);
  pair<int, int> qans =
      make_pair((int)((2LL * (q1.first + q2.first) - q1.second + mod) % mod),
                (int)((q1.second + q2.second) % mod));
  return qans;
}
int main() {
  long long n;
  cin >> n;
  int m;
  scanf("%d%d", &m, &mod);
  for (int i = 0; i < m; ++i) {
    long long l, r, u, v;
    cin >> l >> r >> u >> v;
    pair<int, int> q = query(n, l, r, u, v, 1, n);
    printf("%d\n", q.first);
  }
  return 0;
}