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;
pair<long long, long long> inrange_sum(long long l, long long r, long long u,
                                       long long v, long long n) {
  if (u > v)
    return make_pair(0LL, 0LL);
  else if (l == 1 && r == n) {
    long long cnt = v - u + 1;
    long long mul = u + v;
    if (cnt & 1)
      mul >>= 1;
    else
      cnt >>= 1;
    long long sum = ((cnt % mod) * (mul % mod)) % mod;
    return make_pair(sum, v - u + 1);
  }
  long long oddcnt = (n + 1LL) >> 1;
  pair<long long, long long> osum, esum;
  long long sum = 0LL;
  if (oddcnt >= l) {
    osum = inrange_sum(l, min(r, oddcnt), ((u | 1LL) + 1LL) >> 1,
                       (v + 1LL) >> 1, oddcnt);
    sum = (osum.first << 1) - osum.second;
  }
  if (oddcnt < r) {
    esum = inrange_sum(max(1LL, l - oddcnt), r - oddcnt, (u + 1LL) >> 1, v >> 1,
                       n >> 1);
    sum += esum.first << 1;
  }
  sum %= mod;
  if (sum < 0) sum += mod;
  return make_pair(sum, osum.second + esum.second);
}
int main() {
  long long n;
  int m;
  scanf("%I64d%d%I64d", &n, &m, &mod);
  long long l, r, u, v;
  for (int i = 0; i < m; ++i) {
    scanf("%I64d%I64d%I64d%I64d", &l, &r, &u, &v);
    v = min(v, n);
    pair<long long, long long> sum = inrange_sum(l, r, u, v, n);
    printf("%I64d\n", sum.first);
  }
  return 0;
}