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:60000000")
using namespace std;
template <class T>
T sqr(T x) {
  return x * x;
}
double const pi = 3.1415926535897932384626433832795;
int const inf = (int)1e9;
long long const inf64 = (long long)4e18;
const string name = "d";
long long P, n, ans, A, B, L, R;
int m;
inline long long calc(long long zn, long long p, long long lvl) {
  long long max = (lvl - zn) / p * p + zn;
  if (max + p <= lvl) max += p;
  if (max > lvl) max -= p;
  long long X = 1 + (max - zn) / p;
  long long Y = max + zn;
  if ((X & 1) == 0)
    X >>= 1;
  else
    Y >>= 1;
  long long res = ((X % P) * (Y % P)) % P;
  return res;
}
void sum(long long l, long long r, long long zn, long long p) {
  if (L > r || l > R) return;
  if (L <= l && r <= R) {
    if (zn == p) zn = 0;
    ans += calc(zn, p, B) - calc(zn, p, A);
    if (ans < 0) ans += P;
    if (ans >= P) ans -= P;
    return;
  }
  sum(l, (l + r) >> 1, zn, 2 * p);
  sum((l + r + 2) >> 1, r, zn + p, 2 * p);
}
int main() {
  cin >> n >> m >> P;
  for (int i = 0; i < (int)m; i++) {
    cin >> L >> R >> A >> B;
    if (A > n)
      printf("0\n");
    else {
      B = min(n, B);
      A--;
      ans = 0;
      sum(1, n, 1, 1);
      printf("%d\n", ans);
    }
  }
  cerr << clock() << endl;
  return 0;
}