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;
long long Mod, LB, UB;
long long GetNumber() {
  char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  long long X = c - 48;
  while (c = getchar(), c >= '0' && c <= '9') X = X * 10 + c - 48;
  return X;
}
void Output(long long X) {
  int P = 0, A[20];
  while (X) {
    A[P++] = X % 10;
    X /= 10;
  }
  if (!P) A[P++] = 0;
  while (P--) putchar(A[P] + 48);
  putchar('\n');
}
long long Query(int Depth, long long Step, long long P, long long Q,
                long long Left, long long Right, long long L, long long R) {
  if (LB > Q || UB < P) return 0;
  if (L == Left && R == Right) {
    long long Min = 0, Max = (Q - P) >> Depth, T;
    if (LB > P) Min = ((LB - P - 1) >> Depth) + 1;
    if (UB < Q) Max = (UB - P) >> Depth;
    if (Min > Max) return 0;
    long long Count = Max - Min + 1, S1 = (Count % Mod) * (P % Mod) % Mod,
              X = Min + Max;
    if (!(X & 1))
      X /= 2;
    else
      Count /= 2;
    Count %= Mod;
    X %= Mod;
    long long S2 = Count * X % Mod * (Step % Mod) % Mod;
    return (S1 + S2) % Mod;
  }
  long long Mid = (Left + Right) / 2;
  if ((Q - P) & Step) {
    if (R <= Mid)
      return Query(Depth + 1, Step << 1, P, Q - Step, Left, Mid, L, R);
    if (L > Mid)
      return Query(Depth + 1, Step << 1, P + Step, Q, Mid + 1, Right, L, R);
    return (Query(Depth + 1, Step << 1, P, Q - Step, Left, Mid, L, Mid) +
            Query(Depth + 1, Step << 1, P + Step, Q, Mid + 1, Right, Mid + 1,
                  R)) %
           Mod;
  } else {
    if (R <= Mid) return Query(Depth + 1, Step << 1, P, Q, Left, Mid, L, R);
    if (L > Mid)
      return Query(Depth + 1, Step << 1, P + Step, Q - Step, Mid + 1, Right, L,
                   R);
    return (Query(Depth + 1, Step << 1, P, Q, Left, Mid, L, Mid) +
            Query(Depth + 1, Step << 1, P + Step, Q - Step, Mid + 1, Right,
                  Mid + 1, R)) %
           Mod;
  }
}
int main() {
  long long N = GetNumber(), M = GetNumber();
  Mod = GetNumber();
  while (M--) {
    long long L = GetNumber(), R = GetNumber();
    LB = GetNumber();
    UB = GetNumber();
    Output(Query(0, 1, 1, N, 1, N, L, R));
  }
  return 0;
}