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 p[100];
long long u, v;
int MOD;
void addmod(int &x, int y) { x = (x + y) % MOD; }
int dfs(long long minx, long long maxx, long long left, long long right,
        long long state, int step) {
  if (left <= minx && maxx <= right) {
    int ret = 0;
    long long r = (v >= state) ? ((v - state) >> step) : -1;
    long long l = (u - 1 >= state) ? ((u - 1 - state) >> step) : -1;
    if (r < 0LL) return 0;
    long long a = l + r + 1, b = r - l;
    if (a & 1)
      b >>= 1;
    else
      a >>= 1;
    ret = a % MOD * (b % MOD) % MOD;
    ret = (long long)ret * p[step] % MOD;
    addmod(ret, (r - l) % MOD * (state % MOD) % MOD);
    addmod(ret, (r - l) % MOD);
    return ret;
  }
  long long mid = (minx + maxx) / 2;
  int ret = 0;
  if (left <= mid) addmod(ret, dfs(minx, mid, left, right, state, step + 1));
  if (right > mid)
    addmod(ret,
           dfs(mid + 1, maxx, left, right, state | (1LL << step), step + 1));
  return ret;
}
int main() {
  long long n, l, r;
  int m;
  scanf("%I64d%d%d", &n, &m, &MOD);
  p[0] = 1 % MOD;
  for (int i = 1; i <= 90; ++i) {
    p[i] = (p[i - 1] * 2) % MOD;
  }
  for (int i = 0; i < m; ++i) {
    scanf("%I64d%I64d%I64d%I64d", &l, &r, &u, &v);
    u = max(0LL, u - 1);
    v = min(n - 1, v - 1);
    if (u > v)
      puts("0");
    else
      printf("%d\n", dfs(0, n - 1, l - 1, r - 1, 0, 0));
  }
  return 0;
}