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;
static uint_fast32_t mod;
struct Residue {
  uint_fast32_t n;
};
static inline Residue operator+(const Residue &x, const Residue &y) noexcept {
  return {(x.n + y.n) % mod};
}
static inline Residue operator-(const Residue &x, const Residue &y) noexcept {
  return {(x.n + mod - y.n) % mod};
}
static inline Residue operator*(const Residue &x, const Residue &y) noexcept {
  return {(static_cast<uint_fast64_t>(x.n) * static_cast<uint_fast64_t>(y.n)) %
          mod};
}
static inline Residue operator<<(Residue x, uint_fast8_t k) noexcept {
  for (; k; k--) {
    x.n = (x.n << 1u) % mod;
  }
  return x;
}
static inline ostream &operator<<(ostream &os, const Residue &x) noexcept {
  return os << x.n;
}
struct S {
  uint_fast64_t a;
  Residue s;
};
static inline S operator+(const S &x, const S &y) noexcept {
  return {x.a + y.a, x.s + y.s};
}
static inline S operator-(const S &x, const S &y) noexcept {
  return {x.a - y.a, x.s - y.s};
}
static S g(const uint_fast8_t k, const uint_fast64_t b,
           const uint_fast64_t w) noexcept {
  if (w <= b) {
    return {0, Residue{0}};
  } else {
    const uint_fast64_t m = (w - b - 1) >> k;
    const Residue z = m & 1u ? Residue{m % mod} * Residue{((m + 1) / 2) % mod}
                             : Residue{(m / 2) % mod} * Residue{(m + 1) % mod};
    return {m + 1, (z << k) + Residue{(m + 1) % mod} * Residue{b % mod}};
  }
}
static uint_fast64_t u, v;
static S f(const uint_fast64_t s, const uint_fast64_t t, const uint_fast8_t k,
           const uint_fast64_t b, const uint_fast64_t l,
           const uint_fast64_t r) noexcept {
  if (l == s && r == t) {
    return g(k, b, v) - g(k, b, u);
  }
  const uint_fast64_t m = s + (t - s + 1) / 2;
  if (r <= m) return f(s, m, k + 1, 0 + b, l, r);
  if (l >= m)
    return f(m, t, k + 1, (static_cast<uint_fast64_t>(1) << k) + b, l, r);
  return f(s, m, k + 1, 0 + b, l, m) +
         f(m, t, k + 1, (static_cast<uint_fast64_t>(1) << k) + b, m, r);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  uint_fast64_t n;
  uint_fast32_t m;
  cin >> n >> m >> mod;
  for (; m; m--) {
    uint_fast64_t l, r;
    cin >> l >> r >> u >> v;
    u = u - 1;
    v = v - 1 + 1;
    if (u >= n) {
      cout << 0 << '\n';
      continue;
    }
    if (v > n) {
      v = n;
    }
    const S s = f(0, n, 0, 0, l - 1, r - 1 + 1);
    cout << s.s + Residue{s.a % mod} << '\n';
  }
}