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 mi[1000], mo;
long long read() {
  long long ans = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') {
    ans = ans * (long long)10 + (long long)(ch - '0');
    ch = getchar();
  }
  return ans;
}
long long query(long long l, long long r, long long x, long long y, long long k,
                long long head, int lay) {
  if (l == r) {
    if (head <= k) return head;
    return 0;
  }
  if (l == x && y == r) {
    if (head > k) return 0;
    long long x1 = r - l + 1LL;
    long long x2 = ((k - head) / mi[lay - 1]) + 1LL;
    x1 = min(x1, x2);
    if (!x1) return 0;
    long long fir = head, las = head + (x1 - 1LL) * mi[lay - 1];
    fir += las;
    if (fir & 1)
      x1 >>= 1;
    else
      fir >>= 1;
    long long sum1 = ((fir % mo) * (x1 % mo)) % mo;
    return sum1;
  }
  long long mid = (l + r) >> 1;
  if (mid < x) return query(mid + 1, r, x, y, k, head + mi[lay - 1], lay + 1);
  if (mid >= y) return query(l, mid, x, y, k, head, lay + 1);
  return (query(l, mid, x, mid, k, head, lay + 1) +
          query(mid + 1, r, mid + 1, y, k, head + mi[lay - 1], lay + 1)) %
         mo;
}
int main() {
  long long n, m;
  scanf("%lld%lld%lld", &n, &m, &mo);
  mi[0] = 1LL;
  for (int i = 1; i <= 62; i++) {
    mi[i] = mi[i - 1] * 2LL;
  }
  while (m--) {
    long long l = read(), r = read(), u = read(), v = read();
    if (l > r || u > v) {
      puts("0");
      continue;
    }
    long long ans = query(1LL, n, l, r, v, 1LL, 1);
    ans -= query(1LL, n, l, r, u - 1LL, 1LL, 1);
    ans = (ans % mo + mo) % mo;
    printf("%lld\n", ans);
  }
  return 0;
}