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 int mod;
long long int n;
inline long long int getMod(long long int v) { return ((v % mod + mod)) % mod; }
long long int seriesSum(long long int a, long long int d, long long int n) {
  long long int ret = 0;
  assert(d % 2 == 0);
  d /= 2;
  a %= mod;
  n %= mod;
  ret = getMod((a + getMod(n - 1) * d));
  ret = (n * ret) % mod;
  return ret;
}
long long int oddSeries(long long int a, long long int n, long long int d,
                        long long int u, long long int x) {
  if (a > u) return 0;
  long long int tm = (u - a) / d + 1;
  long long int mn = min(n, tm);
  return (seriesSum(a, d, mn) + (getMod(x) * getMod(mn)) % mod) % mod;
}
long long int evenSeries(long long int a, long long int n, long long int d,
                         long long int u, long long int x) {
  if (a > u) return 0;
  long long int tm = (u - a) / d + 1;
  long long int mn = min(n, tm);
  return (seriesSum(a, d, mn) + (getMod(x) * getMod(mn)) % mod) % mod;
}
long long int oddCumulative(long long int size, long long int gap,
                            long long int u, long long int l,
                            long long int constant) {
  if (u < 0)
    return 0;
  else if (l <= 0) {
    return 0;
  } else if (size <= l) {
    return oddSeries(1, size, gap, u, constant);
  } else {
    long long int leftSize = size / 2 + size % 2;
    long long int rightSize = size - leftSize;
    return (oddCumulative(leftSize, gap * 2, u, l, constant) +
            (oddCumulative(rightSize, gap * 2, u - gap, l - leftSize,
                           (constant + gap) % mod))) %
           mod;
  }
}
long long int evenCumulative(long long int size, long long int gap,
                             long long int u, long long int l,
                             long long int constant) {
  if (u < 0)
    return 0;
  else if (l <= 0) {
    return 0;
  } else if (size <= l) {
    return evenSeries(2, size, gap, u, constant);
  } else {
    long long int leftSize = size / 2 + size % 2;
    long long int rightSize = size - leftSize;
    return (evenCumulative(leftSize, gap * 2, u, l, constant) +
            (evenCumulative(rightSize, gap * 2, u - gap, l - leftSize,
                            (constant + gap) % mod))) %
           mod;
  }
}
long long int query(long long int l, long long int u) {
  long long int leftSize = n / 2 + n % 2;
  long long int rightSize = n - leftSize;
  return oddCumulative(leftSize, 2, u, l, 0) +
         evenCumulative(rightSize, 2, u, l - leftSize, 0);
}
long long int original[500];
long long int manualQuery(long long int x, long long int y, long long int a,
                          long long int b) {
  long long int ret = 0;
  int i, j;
  for (i = x; i <= y; i++) {
    if (a <= original[i] && original[i] <= b) {
      ret += original[i];
      ret %= mod;
    }
  }
  return ret;
}
int main() {
  int m;
  scanf("%lld %d %lld", &n, &m, &mod);
  int i, j;
  for (i = 0; i < m; i++) {
    long long int l, r, u, v;
    scanf("%lld %lld %lld %lld", &l, &r, &u, &v);
    long long int sum = getMod((query(r, v) - query(r, u - 1)));
    sum = getMod(sum - getMod((query(l - 1, v) - query(l - 1, u - 1))));
    printf("%lld\n", sum);
  }
  return 0;
}