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 n, m, p, ans, ql, qr, qb, qt;
void get(long long minnum, long long deg, long long len) {
  long long maxnum = minnum + deg * (len - 1);
  if (qt < minnum) return;
  if (qb > maxnum) return;
  long long s1, s2;
  if (qb <= minnum)
    s1 = minnum;
  else if ((qb - minnum) % deg == 0)
    s1 = qb;
  else
    s1 = ((qb - minnum) / deg + 1) * deg + minnum;
  if (qt >= maxnum)
    s2 = maxnum;
  else if ((maxnum - qt) % deg == 0)
    s2 = qt;
  else
    s2 = maxnum - ((maxnum - qt) / deg + 1) * deg;
  long long nn = (s2 - s1) / deg;
  long long nn1 = nn + 1;
  s1 = s1 % p;
  s1 = (s1 * (nn1 % p)) % p;
  if (nn & 1)
    nn1 >>= 1;
  else
    nn >>= 1;
  nn %= p;
  nn1 %= p;
  nn = (nn * nn1) % p;
  deg = deg % p;
  deg = (deg * nn) % p;
  ans = (ans + s1 + deg) % p;
}
void f(long long li, long long minnum, long long deg, long long len) {
  long long ri = li + len - 1;
  if (ql <= li && qr >= ri) {
    get(minnum, deg, len);
    return;
  }
  long long ll = (len + 1) >> 1;
  long long mi = li + ll - 1;
  if (ql <= mi) f(li, minnum, deg << 1, ll);
  if (qr > mi) f(li + ll, minnum + deg, deg << 1, len - ll);
}
void solve() {
  ans = 0;
  f(1, 1, 1, n);
  cout << ans << "\n";
}
int main() {
  cin >> n >> m >> p;
  while (m--) {
    cin >> ql >> qr >> qb >> qt;
    solve();
  }
  return 0;
}