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;
int m;
long long n, ans, mo, L, R, x, y;
long long Sum(long long S, long long d, long long num) {
  long long xx, yy, res;
  S %= mo;
  d %= mo;
  xx = S * (num % mo) % mo;
  if (num & 1)
    yy = (num % mo * ((num - 1) / 2 % mo) % mo) * d % mo;
  else
    yy = (num / 2 % mo * ((num - 1) % mo) % mo) * d % mo;
  res = (xx + yy) % mo;
  return res;
}
long long Calc(long long l, long long r, long long x, long long y, long long s,
               long long d, long long before) {
  long long res = 0;
  if (l == r) {
    if (before + l >= L && before + l <= R) res = s % mo;
    return res;
  }
  long long mid = (l + r) / 2;
  long long X1, Y1, X2, Y2, S1, S2, D;
  X1 = x / 2 + 1;
  Y1 = (y + 1) / 2;
  X2 = (x + 1) / 2;
  Y2 = y / 2;
  if (x & 1)
    S1 = s, S2 = (s + d) % mo;
  else
    S1 = (s + d) % mo, S2 = s;
  D = 2 * d % mo;
  if (X1 <= Y1) {
    if (l >= L && mid <= R)
      res = (res + Sum(S1, D, Y1 - X1 + 1)) % mo;
    else if (l <= R && mid >= L)
      res = (res + Calc(l, mid, X1, Y1, S1, D, before)) % mo;
  }
  if (X2 <= Y2) {
    if (mid + 1 >= L && r <= R)
      res = (res + Sum(S2, D, Y2 - X2 + 1)) % mo;
    else if (mid + 1 <= R && r >= L)
      res = (res + Calc(mid + 1, r, X2, Y2, S2, D, before + mid - l + 1)) % mo;
  }
  return res;
}
int main() {
  scanf("%I64d%d%I64d", &n, &m, &mo);
  for (int i = 1; i <= m; ++i) {
    scanf("%I64d%I64d%I64d%I64d", &L, &R, &x, &y);
    if (y > n) y = n;
    if (x < 1) x = 1;
    if (y < x) {
      printf("0\n");
      continue;
    }
    long long ans = Calc(1, n, x, y, x % mo, 1, 0);
    printf("%I64d\n", ans);
  }
  return 0;
}