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;
long long get(long long n, long long m, long long r) {
  if (n == 0) return 0;
  if (n % 2 == 0) {
    m %= mod;
    m *= ((n - 1) % mod);
    m %= mod;
    m += (2LL * r) % mod;
    m %= mod;
    n /= 2;
    n %= mod;
    return (n * m) % mod;
  } else {
    long long tmp = n;
    tmp %= mod;
    n = (n - 1) / 2;
    n %= mod;
    n *= (m % mod);
    n %= mod;
    n += (r % mod);
    n %= mod;
    return (n * tmp) % mod;
  }
}
pair<long long, long long> func(long long n, long long len, long long m,
                                long long r, long long a, long long b) {
  if (len == 0) return {0, 0};
  if (n == 1 && len == 1) {
    if (r <= b && r > a)
      return {(r) % mod, 1};
    else
      return {0, 0};
  }
  if ((n + 1) / 2 >= len) return func((n + 1) / 2, len, 2LL * m, r, a, b);
  long long cntr, cntl, cnt = 0;
  if (b < r)
    cntr = 0;
  else
    cntr = min((n + 1) / 2, (b - r) / (2LL * m) + 1);
  if (a < r)
    cntl = 0;
  else
    cntl = min((n + 1) / 2, (a - r) / (2LL * m) + 1);
  cnt += cntr - cntl;
  cnt %= mod;
  long long ans = get(cntr, 2LL * m, r) + mod - get(cntl, 2LL * m, r);
  ans %= mod;
  pair<long long, long long> X =
      func(n / 2, len - (n + 1) / 2, 2LL * m, m + r, a, b);
  ans += X.first;
  ans %= mod;
  cnt += X.second;
  cnt %= mod;
  return {ans, cnt};
}
int main() {
  long long n, q;
  cin >> n >> q >> mod;
  for (int i = 0; i < q; i++) {
    long long l, r, u, v;
    cin >> l >> r >> u >> v;
    v--;
    u--;
    pair<long long, long long> X1 = func(n, r, 1, 0, u - 1, v);
    pair<long long, long long> X2 = func(n, l - 1, 1, 0, u - 1, v);
    long long ans = X1.first + X1.second + 2LL * mod - X2.first - X2.second;
    ans %= mod;
    cout << ans << '\n';
  }
}