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;
template <class T>
inline T checkmin(T &a, T b) {
  return (a < b) ? a : a = b;
}
template <class T>
inline T checkmax(T &a, T b) {
  return (a > b) ? a : a = b;
}
template <class T>
T GCD(T a, T b) {
  if (a < 0) return GCD(-a, b);
  if (b < 0) return GCD(a, -b);
  return (a == 0) ? b : GCD(b % a, a);
}
template <class T>
T LCM(T a, T b) {
  if (a < 0) return LCM(-a, b);
  if (b < 0) return LCM(a, -b);
  return (a == 0 || b == 0) ? 0 : a / GCD(a, b) * b;
}
namespace Poor {
long long N, Odd, Even;
int M, Mod;
long long DoOdd(long long L, long long R, long long U, long long V) {
  if (L > R) return 0;
  checkmax(L, U / 2LL + 1);
  checkmin(R, (V + 1LL) / 2LL);
}
long long DoEven(long long L, long long R, long long U, long long V) {
  if (L > R) return 0;
}
void Run() {
  cin >> N >> M >> Mod;
  Odd = (N + 1) / 2LL;
  Even = N / 2LL;
  for (int i = 0; i < M; ++i) {
    long long L, R, U, V;
    cin >> L >> R >> U >> V;
    cout << (DoOdd(L, min(R, Odd), U, V) + DoEven(max(Odd + 1, L), R, U, V)) %
                Mod
         << endl;
  }
}
}  // namespace Poor
namespace Poor2 {
long long N, U, V, Mod;
int M;
long long Ask(long long a1, long long d, long long n, long long S,
              long long T) {
  if (S == 1 && T == n) {
    long long Left = max((U - a1 + d - 1LL) / d, 0LL);
    long long Right = min((V - a1) / d, n - 1LL);
    while (U > a1 + Left * d) ++Left;
    while (V < a1 + Right * d) --Right;
    if (Left > Right) return 0;
    return (((((a1 % (Mod * 2LL)) +
               ((Left % (Mod * 2LL)) * (d % (Mod * 2LL))) % (Mod * 2LL) +
               (a1 % (Mod * 2LL)) +
               (Right % (Mod * 2LL) * (d % (Mod * 2LL))) % (Mod * 2LL)) %
              (Mod * 2LL))) *
            ((Right - Left + 1LL) % (Mod * 2LL)) % (Mod * 2LL)) /
           2LL;
  }
  long long odd = (n + 1LL) / 2LL;
  long long res = 0;
  if (S <= odd) res += Ask(a1, d * 2LL, odd, S, min(T, odd));
  if (odd < T)
    res += Ask(a1 + d, d * 2LL, n - odd, max(S, odd + 1LL) - odd, T - odd);
  res %= Mod;
  return res;
}
void Run() {
  cin >> N >> M >> Mod;
  for (int i = 0; i < M; ++i) {
    long long S, T;
    cin >> S >> T >> U >> V;
    cout << Ask(1, 1, N, S, T) << endl;
  }
}
}  // namespace Poor2
int main(int argc, char **argv) {
  ios::sync_with_stdio(false);
  Poor2::Run();
}