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 JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.math.*;

public class Main implements Runnable {
    int mod;

    long tmpr, tmps;
    
    void calc(long l, long r, long u, long v, long n) {
        u = Math.max(u, 1);
        v = Math.min(v, n);

        if (u > v) { tmpr = tmps = 0; return; }
        if (l==1 && r==n) {
            long a1 = u+v, a2 = (v-u+1); if (a1%2 == 0) a1 /= 2; else a2 /= 2;
            tmpr = ((a1%mod) * (a2%mod))%mod;
            tmps = v-u+1;
            return;
        }
        
        long m = (n+1)/2;
        long resr = 0, ress = 0;
        
        if (l <= m) { calc(l, Math.min(m, r), u/2+1, (v+1)/2, m); resr += 2*tmpr-tmps; ress += tmps; }
        if (r > m) { calc(Math.max(1, l-m), r-m, (u+1)/2, v/2, n-m); resr += 2*tmpr; ress += tmps; }
        
        tmpr = resr%mod; if (tmpr < 0) tmpr += mod; tmps = ress;
    }
    
    public void run() {
        long n = nextLong();
        int m = nextInt();
        mod = nextInt();
        
        for (int i = 0; i < m; ++i) {
            long l = nextLong(), r = nextLong(), u = nextLong(), v = nextLong();
            
            calc(l, r, u, v, n);
            output.println(tmpr);
        }

        output.flush();
    }

    public static void main(String[] args) {
        //try { input = new BufferedReader(new InputStreamReader(new FileInputStream("input.txt"))); } catch (Exception e) {}
        new Main().run();
    }

    static BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter output = new PrintWriter(System.out);
    StringTokenizer tokenizer = null;

    int nextInt() {
        return Integer.parseInt(nextToken());
    }

    long nextLong() {
        return Long.parseLong(nextToken());
    }

    double nextDouble() {
        return Double.parseDouble(nextToken());
    }

    String nextToken() {
        try
        {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                tokenizer = new StringTokenizer(input.readLine());
            }

            return tokenizer.nextToken();
        } catch (Exception e) {
            return null;
        }
    }
}