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>
long long n;
int m, mod;
long long u, v;
inline long long min(long long a, long long b) { return a < b ? a : b; }
inline long long max(long long a, long long b) { return a > b ? a : b; }
inline long long solve(long long q, long long w, long long l, long long r,
                       long long beg, long long st) {
  if (q == l && w == r) {
    long long n = (r - l + 1);
    long long las = beg + (n - 1) * st;
    if (u < beg || v > las) return 0;
    long long vv = (v - beg + st - 1) / st;
    long long uu = (u - beg) / st;
    vv = max(vv, 0);
    uu = min(uu, n - 1);
    if (vv > uu) return 0;
    long long f = beg + vv * st;
    long long g = beg + uu * st;
    long long cnt = (uu - vv + 1);
    f %= 2 * mod;
    g %= 2 * mod;
    cnt %= 2 * mod;
    return (((f + g) % (2 * mod) * cnt) / 2) % mod;
  }
  long long oddc = (q + w) / 2;
  long long ans = 0;
  if (r <= oddc)
    ans = (ans + solve(q, oddc, l, r, beg, st * 2)) % mod;
  else if (l <= oddc)
    ans = (ans + solve(q, oddc, l, oddc, beg, st * 2)) % mod;
  if (l > oddc)
    ans = (ans + solve(oddc + 1, w, l, r, beg + st, st * 2)) % mod;
  else if (r > oddc)
    ans = (ans + solve(oddc + 1, w, oddc + 1, r, beg + st, st * 2)) % mod;
  return ans;
}
int main() {
  scanf("%I64d%d%d", &n, &m, &mod);
  while (m--) {
    long long l, r;
    scanf("%I64d%I64d%I64d%I64d", &l, &r, &v, &u);
    printf("%d\n", (int)solve(1, n, l, r, 1, 1));
  }
}