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>
inline long long R() {
  char c;
  long long sign = 1, res = 0;
  while ((c = getchar()) > '9' || c < '0')
    if (c == '-') sign = -1;
  res += c - '0';
  while ((c = getchar()) >= '0' && c <= '9') res = res * 10 + c - '0';
  return res * sign;
}
long long n, m, Mo;
long long solve(long long l, long long r, long long z, long long y,
                long long Mi, long long Ma, long long deep, long long sta,
                long long en) {
  long long res = 0;
  if (l > y || r < z) return 0;
  if (l >= z && r <= y) {
    if (Mi > en || Ma < sta) return 0;
    long long p1 = en % deep;
    if (Ma < en) {
      if (Ma / deep * deep + p1 > Ma)
        Ma = Ma / deep * deep + p1 - deep;
      else
        Ma = Ma / deep * deep + p1;
    } else
      Ma = en;
    if (Mi > sta) {
      if (Mi / deep * deep + p1 >= Mi)
        Mi = Mi / deep * deep + p1;
      else
        Mi = Mi / deep * deep + p1 + deep;
    } else
      Mi = sta;
    if (Mi > Ma) return 0;
    if ((Mi + Ma) % 2 == 0) {
      res = ((Mi + Ma >> 1) % Mo) * ((1 + (Ma - Mi) / deep) % Mo);
    } else {
      res = ((Mi + Ma) % Mo) * (((1 + (Ma - Mi) / deep) / 2) % Mo);
    }
    res = res % Mo;
    return res;
  }
  res += solve(l, (l + r >> 1), z, y, Mi, Ma, deep * 2, sta,
               (r - l + 1) & 1 ? en : en - deep);
  res += solve((l + r >> 1) + 1, r, z, y, Mi, Ma, deep * 2, sta + deep,
               (r - l + 1) & 1 ? en - deep : en);
  return res % Mo;
}
signed main() {
  long long l, r, Mi, Ma;
  n = R();
  m = R();
  Mo = R();
  while (m--) {
    l = R();
    r = R();
    Mi = R();
    Ma = R();
    printf("%lld\n", (long long)solve(1, n, l, r, Mi, Ma, 1, 1, n));
  }
}