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>
#pragma optimize(2)
using namespace std;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  bool f = false;
  for (; !isdigit(c); c = getchar()) f != c == '-';
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  if (f) x = -x;
}
template <typename T, typename... Arg>
inline void read(T &x, Arg &...args) {
  read(x);
  read(args...);
}
template <typename T>
inline void write(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
long long n, m, mod;
long long get(long long fs, long long num, long long lim, long long d) {
  if (lim < fs) return 0;
  lim = (lim - fs) / d * d + fs;
  long long ls = min(lim, fs + d * (num - 1));
  long long v1 = (ls - fs) / d + 1, v2 = (fs + ls);
  if (v1 % 2 == 0)
    v1 /= 2;
  else
    v2 /= 2;
  return ((v1) % mod) * ((v2) % mod) % mod;
}
long long solve(long long lim, long long fs, long long d, long long num,
                long long len) {
  if (num == 0) return 0;
  if (len == 1) return fs <= lim ? fs : 0;
  long long mid = (len + 1) / 2;
  if (num >= mid)
    return ((get(fs, mid, lim, d * 2)) +
            (solve(lim, fs + d, d * 2, num - mid, len - mid))) %
           mod;
  else
    return solve(lim, fs, d * 2, num, mid);
}
signed main() {
  read(n, m, mod);
  while (m--) {
    long long l, r, x, y;
    read(l, r, x, y);
    write((((solve(y, 1, 1, r, n) - solve(x - 1, 1, 1, r, n) + mod) % mod -
            (solve(y, 1, 1, l - 1, n) - solve(x - 1, 1, 1, l - 1, n) + mod) %
                mod) +
           mod) %
          mod);
    putchar('\n');
  }
}