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;
template <class T>
void pv(T a, T b) {
  for (T i = a; i != b; ++i) cout << *i << " ";
  cout << endl;
}
template <class T>
void pvp(T a, T b) {
  for (T i = a; i != b; ++i)
    cout << "(" << i->first << ", " << i->second << ") ";
  cout << endl;
}
template <class T>
void chmin(T &t, T f) {
  if (t > f) t = f;
}
template <class T>
void chmax(T &t, T f) {
  if (t < f) t = f;
}
int in_c() {
  int c;
  for (; (c = getchar()) <= ' ';) {
    if (!~c) throw ~0;
  }
  return c;
}
int in() {
  int x = 0, c;
  for (; (unsigned)((c = getchar()) - '0') >= 10;) {
    if (c == '-') return -in();
    if (!~c) throw ~0;
  }
  do {
    x = (x << 3) + (x << 1) + (c - '0');
  } while ((unsigned)((c = getchar()) - '0') < 10);
  return x;
}
long long In() {
  long long x = 0, c;
  for (; (unsigned)((c = getchar()) - '0') >= 10;) {
    if (c == '-') return -In();
    if (!~c) throw ~0;
  }
  do {
    x = (x << 3) + (x << 1) + (c - '0');
  } while ((unsigned)((c = getchar()) - '0') < 10);
  return x;
}
void Out(long long x) {
  if (x >= 10) Out(x / 10);
  putchar('0' + x % 10);
}
long long N;
int Q;
long long MO;
long long mul(long long a, long long b) {
  a %= MO;
  b %= MO;
  return a * b % MO;
}
long long mulDiv2(long long a, long long b) {
  a %= MO;
  b %= MO;
  if (a % 2 == 0)
    a /= 2;
  else
    b /= 2;
  return a * b % MO;
}
long long calc(long long u, long long p, long long m) {
  chmin(u, N);
  if (u < p) return 0;
  long long k = (u - p) / m;
  return (mul(mulDiv2(k, k + 1), m) + mul(k + 1, p)) % MO;
}
long long calc(long long u, long long v, long long p, long long m) {
  long long ret = 0;
  ret += calc(v, p, m);
  ret -= calc(u - 1, p, m);
  ret %= MO;
  return ret;
}
long long solve(long long l, long long r, long long u, long long v, long long a,
                long long b, long long p, long long m) {
  chmax(l, a);
  chmin(r, b);
  if (l > r) return 0;
  long long ret = 0;
  if (l == a && r == b) {
    ret += calc(u, v, p, m);
  } else {
    long long odd = (b - a + 1 + 1) / 2;
    ret += solve(l, r, u, v, a, a + odd - 1, p, m * 2);
    ret += solve(l, r, u, v, a + odd, b, p + m, m * 2);
  }
  ret %= MO;
  return ret;
}
int main() {
  int q;
  long long l, r, u, v;
  for (; cin >> N >> Q >> MO;) {
    for (q = 0; q < Q; ++q) {
      l = In();
      r = In();
      u = In();
      v = In();
      long long res = solve(l, r, u, v, 1, N, 1, 1);
      res %= MO;
      res += MO;
      res %= MO;
      Out(res);
      puts("");
    }
  }
  return 0;
}