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>
template <typename _T>
void read(_T &x) {
  x = 0;
  char s = getchar();
  int f = 1;
  while (s < '0' || '9' < s) {
    f = 1;
    if (s == '-') f = -1;
    s = getchar();
  }
  while ('0' <= s && s <= '9') {
    x = (x << 3) + (x << 1) + (s - '0'), s = getchar();
  }
  x *= f;
}
template <typename _T>
void write(_T x) {
  if (x < 0) putchar('-'), x = -x;
  if (9 < x) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename _T>
_T MAX(const _T a, const _T b) {
  return a > b ? a : b;
}
template <typename _T>
_T MIN(const _T a, const _T b) {
  return a < b ? a : b;
}
int mod;
int Sub(int x, int v) { return x < v ? x + mod - v : x - v; }
int Mul(long long x, int v) {
  x *= v;
  if (x >= mod) x %= mod;
  return x;
}
int Add(int x, int v) { return x + v >= mod ? x + v - mod : x + v; }
struct Value {
  int s, c;
  Value() { s = c = 0; }
  Value(int S, int C) { s = S, c = C; }
  Value operator+(const Value &b) const {
    return Value(Add(s, b.s), Add(c, b.c));
  }
  Value operator+(const int &b) const { return Value(Add(s, Mul(b, c)), c); }
  Value operator*(const int &b) const { return Value(Mul(s, b), c); }
  Value operator-(const int &b) const { return *this + (mod - b); }
  Value operator-=(const int &b) { return *this = *this - b; }
  Value operator+=(const int &b) { return *this = *this + b; }
  Value operator*=(const int &b) { return *this = *this * b; }
  Value operator+=(const Value &b) { return *this = *this + b; }
};
Value Sum(long long a, long long b) {
  if (a > b) return Value();
  a %= mod, b %= mod;
  int s = Add(a, b), d = Add(1, Sub(b, a));
  if (s & 1) return Value(1ll * s * (d / 2) % mod, d);
  return Value(1ll * (s / 2) * d % mod, d);
}
Value Calc(const long long n, const long long l, const long long r,
           const long long u, const long long v) {
  if (n == 0) return Value();
  if (l <= 1 && n <= r) return Sum(MAX(0ll, u), MIN(n, v));
  long long len = n + 1 >> 1;
  Value ret;
  if (l <= len)
    ret += Calc(len, l, MIN(r, len), u + 2 >> 1, v + 1 >> 1) * 2 - 1;
  if (len < r)
    ret += Calc(n - len, MAX(l - len, 1ll), r - len, u + 1 >> 1, v >> 1) * 2;
  return ret;
}
int main() {
  long long N;
  int M;
  read(N), read(M), read(mod);
  while (M--) {
    long long L, R, U, V;
    read(L), read(R), read(U), read(V);
    write(Calc(N, L, R, U, V).s), putchar('\n');
  }
  return 0;
}