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:64000000")
using namespace std;
const double pi = 3.1415926535897932384626433832795;
double eps = 1e-9;
char ch[2 << 20];
inline string gs() {
  scanf("%s", ch);
  return string(ch);
}
inline string gl() {
  gets(ch);
  return string(ch);
}
inline int gi() {
  int x;
  scanf("%d", &x);
  return x;
}
void error() {
  int ttt = 0;
  cout << 7 / ttt;
}
long long gcd(long long a, long long b) {
  if (!a && !b) return 0;
  if (!b) return a;
  return (!a) ? b : gcd(b % a, a);
}
long long L, U;
long long A, B;
long long mod;
long long go(long long a, long long b, long long n, long long x, long long y) {
  if (A <= x && y <= B) {
    if (a > U) return 0;
    if (a + (n - 1) * b < L) return 0;
    long long f = 0;
    long long e = 0;
    if (a < L) {
      long long o = (L - a + b - 1ll) / b;
      f = a + o * b;
      while (f < L)
        ;
    } else
      f = a;
    if (U >= a + (n - 1ll) * b)
      e = a + (n - 1ll) * b;
    else
      e = a + b * ((U - a) / b);
    if (e < f) return 0;
    long long k = (e - f) / b + 1ll;
    long long r = f + e;
    {
      if (k % 2ll == 0)
        k /= 2LL;
      else {
        while (r % 2ll != 0)
          ;
        r /= 2LL;
      }
      return ((r % mod) * (k % mod)) % mod;
    }
  }
  long long m = (x + y) / 2LL;
  long long res = 0;
  if (A <= m) res = go(a, b * 2ll, (n + 1) / 2LL, x, m);
  if (B > m) res += go(a + b, b * 2ll, n / 2LL, m + 1ll, y);
  if (res >= mod) res -= mod;
  return res;
}
int main() {
  long long n;
  cin >> n;
  int m = gi();
  mod = gi();
  for (int j = (0), _b(m); j < _b; ++j) {
    cin >> A >> B >> L >> U;
    cout << go(1, 1, n, 1, n) << endl;
  }
  return 0;
}