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;
inline long long two(long long x) { return 1LL << x; }
long long n, m, M;
long long mf(long long l, long long r, long long segl, long long segr,
             long long val, long long dep, long long mask) {
  long long mid = segl / 2 + segr / 2 + (segl & segr & 1);
  if (l == segl && r == segr) {
    long long tt = (val >> dep) % M;
    long long t1 = (val >> dep);
    long long t2 = (val >> dep) - 1;
    if (t1 & 1)
      t2 >>= 1;
    else
      t1 >>= 1;
    t1 %= M;
    t2 %= M;
    long long ret = ((t1 * t2 % M) * (two(dep) % M)) % M;
    ret = (ret + tt * (mask % M) + tt) % M;
    if (mask <= (val & (two(dep) - 1)))
      ret = ((ret + mask) % M + (val & (~(two(dep) - 1LL))) + 1LL) % M;
    return ret;
  }
  if (r <= mid)
    return mf(l, r, segl, mid, val, dep + 1, mask);
  else if (l > mid)
    return mf(l, r, mid + 1, segr, val, dep + 1, mask | two(dep));
  else {
    long long t1 = mf(l, mid, segl, mid, val, dep + 1, mask);
    long long t2 = mf(mid + 1, r, mid + 1, segr, val, dep + 1, mask | two(dep));
    return (t1 + t2) % M;
  }
}
int main() {
  while (cin >> n >> m >> M) {
    while (m--) {
      long long l, r, u, v;
      cin >> l >> r >> u >> v;
      l--, r--, u--, v--;
      v = min(n - 1, v);
      long long ret;
      if (u > v)
        ret = 0;
      else if (u)
        ret = mf(l, r, 0, n - 1, v, 0, 0) - mf(l, r, 0, n - 1, u - 1, 0, 0);
      else
        ret = mf(l, r, 0, n - 1, v, 0, 0);
      ret = (ret % M + M) % M;
      cout << ret << endl;
    }
  }
  return 0;
}