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 n, Mod, P[65], u, v;
int m;
long long Query(long long L, long long R, long long l, long long r, int dep,
                long long head) {
  if (L == l && r == R) {
    long long u1 = head, v1 = head + (R - L) * P[dep];
    if (v1 < u || u1 > v) return 0;
    long long u2 = u, v2 = v;
    u2 -= head;
    if (u2 <= 0)
      u2 = 0;
    else if (u2 % P[dep])
      u2 = (u2 - 1) / P[dep] * P[dep] + P[dep];
    u2 += head;
    v2 -= head;
    if (v2 <= 0)
      v2 = 0;
    else if (v2 % P[dep])
      v2 = v2 / P[dep] * P[dep];
    v2 += head;
    u1 = max(u1, u2);
    v1 = min(v1, v2);
    long long r1 = u1 + v1, r2 = (v1 - u1) / P[dep] + 1;
    if (r1 % 2)
      r2 /= 2;
    else
      r1 /= 2;
    return (r1 % Mod) * (r2 % Mod) % Mod;
  }
  long long mid = (L + R) >> 1;
  if (mid >= r)
    return Query(L, mid, l, r, dep + 1, head);
  else if (mid < l)
    return Query(mid + 1, R, l, r, dep + 1, head + P[dep]);
  else
    return (Query(L, mid, l, mid, dep + 1, head) +
            Query(mid + 1, R, mid + 1, r, dep + 1, head + P[dep])) %
           Mod;
}
int main() {
  scanf("%lld%d%lld", &n, &m, &Mod);
  for (int i = 0; i < 65; i++) P[i] = (1ll << i);
  while (m--) {
    long long L, R;
    scanf("%lld%lld%lld%lld", &L, &R, &u, &v);
    printf("%lld\n", Query(1, n, L, R, 0, 1));
  }
  return 0;
}