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 getByIndex(long long n, long long index) {
  long long ret = 0, tpow = 1;
  bool larger = false;
  for (int i = 0; i < 60; i++) {
    bool sub = (larger && n % 2 == 0);
    if (index >= n / 2 + 1 - sub) {
      index -= n / 2 + 1 - sub;
      ret += tpow;
      if (n % 2 == 0) larger = true;
    } else if (n % 2 == 1)
      larger = false;
    n /= 2;
    tpow *= 2;
  }
  return ret;
}
int mod;
struct Modint {
  int val;
  Modint(long long nval) { val = nval % mod; }
  Modint(int nval = 0) { val = nval % mod; }
  Modint& operator+=(Modint r) {
    val = (val + r.val) % mod;
    return *this;
  }
  Modint& operator-=(Modint r) {
    val = (val + mod - r.val) % mod;
    return *this;
  }
  Modint& operator*=(Modint r) {
    val = 1LL * val * r.val % mod;
    return *this;
  }
  Modint operator+(Modint r) { return Modint(*this) += r; }
  Modint operator-(Modint r) { return Modint(*this) -= r; }
  Modint operator*(Modint r) { return Modint(*this) *= r; }
};
long long rev(long long val) {
  long long tpow = 1, ret = 0;
  for (long long i = 59; i >= 0; i--, tpow *= 2)
    ret += ((val & (1LL << i)) != 0) * tpow;
  return ret;
}
Modint getCount(long long lbound, long long rbound, int mod) {
  bool larger = false;
  long long ret = 0, tpow = 1, cnt = 0, rbsum[61], sofar = 0;
  rbsum[60] = 0;
  for (int i = 59; i >= 0; i--) {
    bool one = ((rbound & (1LL << i)) != 0);
    rbsum[i] = (rbsum[i + 1] * 4 + (rbound >> (i + 1)) + 1 -
                !one * (1 + (rbound >> i))) %
               mod;
  }
  for (int i = 0; i < 60; i++) {
    bool sub = (larger && rbound % 2 == 0);
    if (lbound % 2 == 1) {
      cnt += rbound / 2 + 1 - sub;
      ret = (ret + rbsum[i + 1] * tpow * 2 + sofar * (rbound / 2 + 1) -
             (tpow * rbound + sofar) * sub) %
            mod;
      sofar += tpow;
      if (rbound % 2 == 0) larger = true;
    } else if (rbound % 2 == 1)
      larger = false;
    lbound /= 2;
    rbound /= 2;
    tpow = (tpow * 2) % mod;
  }
  cnt += !larger;
  ret += sofar * !larger;
  return ret + cnt + mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n, l, r, u, v;
  int m;
  cin >> n >> m >> mod;
  for (int i = 0; i < m; i++) {
    cin >> l >> r >> u >> v;
    v = min(v, n);
    u = min(u, n + 1);
    r = getByIndex(n - 1, r - 1);
    Modint val = getCount(r, v - 1, mod);
    if (u > 1) val -= getCount(r, u - 2, mod);
    if (l > 1) {
      l = getByIndex(n - 1, l - 2);
      val -= getCount(l, v - 1, mod);
      if (u > 1) val += getCount(l, u - 2, mod);
    }
    cout << val.val << '\n';
  }
  return 0;
}