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 void pisz(int n) { printf("%d\n", n); }
template <typename T, typename TT>
ostream& operator<<(ostream& s, pair<T, TT> t) {
  return s << "(" << t.first << "," << t.second << ")";
}
template <typename T>
ostream& operator<<(ostream& s, vector<T> t) {
  for (int(i) = 0; (i) < (((int)(t.size()))); ++(i)) s << t[i] << " ";
  return s;
}
long long n;
int m, mod;
long long u, v;
long long arazybprzezdwa(long long a, long long b) {
  if (a & 1LL) {
    b /= 2;
  } else {
    a /= 2;
  }
  b %= mod;
  a %= mod;
  return (a * b) % mod;
}
long long go(long long l, long long r, long long le, long long re,
             long long seq, long long pot2) {
  if (l == le && r == re) {
    if (seq == pot2) seq = 0;
    long long u1 = u - seq + pot2;
    if (u1 % pot2 == 0)
      u1 /= pot2;
    else
      u1 = u1 / pot2 + 1;
    long long v1 = v - seq + pot2;
    v1 /= pot2;
    long long d = arazybprzezdwa(u1 + v1, v1 - u1 + 1);
    d = (d * (pot2 % mod)) % mod;
    d += ((seq - pot2) % mod) * ((v1 - u1 + 1) % mod);
    return d % mod;
  }
  long long mid = (le + re) / 2;
  if (r <= mid) return go(l, r, le, mid, seq, pot2 * 2);
  if (l >= mid + 1) return go(l, r, mid + 1, re, seq + pot2, pot2 * 2);
  return (go(l, mid, le, mid, seq, pot2 * 2) +
          go(mid + 1, r, mid + 1, re, seq + pot2, pot2 * 2)) %
         mod;
}
int main() {
  scanf("%I64d %d %d", &n, &m, &mod);
  while (m--) {
    long long l, r;
    scanf("%I64d %I64d %I64d %I64d", &l, &r, &u, &v);
    if (u > n) {
      pisz(0);
      continue;
    }
    if (v > n) v = n;
    long long wyn = go(l, r, 1, n, 1, 1);
    wyn %= mod;
    if (wyn < 0) wyn += mod;
    pisz(wyn);
  }
}