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.*;

public class Main
{
    static long n;
    static int m, mod;
    static long start = System.currentTimeMillis();
    
    static long query(long l, long r, long u, long v, long start, long end, long startvalue, long step)
    {
        long newstart = start, newend = end, newstartvalue = startvalue, newstep = step;
        while (start < l || end > r)
        {
            if (end >= r)
            {
                newstart = start;
                newend = end;
                newstartvalue = startvalue;
                newstep = step;
            }
            long rightstart = start + ((end - start) >> 1) + 1;
            long rightstartvalue = startvalue + step;
            if (rightstart <= l)
            {
                start = rightstart;
                startvalue = rightstartvalue;
            }
            else
            {
                end = rightstart - 1;
            }
            //step *= 2;
            step <<= 1;
        }
        long result;
        if (u > startvalue + step * (end - start) || v < startvalue)
        {
            result = 0;
        }
        else
        {
            long st = (u - startvalue) / step * step + startvalue;
            if (st < u)
            {
                st += step;
            }
            long en = (v - startvalue) / step * step + startvalue;
            if (st < startvalue)
                st = startvalue;
            if (en > startvalue + step * (end - start))
                en = startvalue + step * (end - start);
            
            long sten = st + en;
            long count = (en - st) / step + 1;
            if ((sten & 1) == 0)
                sten >>= 1;
            else
                count >>= 1;
            
            result = (sten % mod) * (count % mod) % mod;
        }
        /*
        for (long i = 0; start + i <= end; i++)
        {
            long inter = startvalue + i * step;
            if (inter >= u && inter <= v)
            {
                result += inter; 
                result %= mod;
            }
        }
        */
        if (end < r)
        {
            result += query(end + 1, r, u, v, newstart, newend, newstartvalue, newstep);
            result %= mod;
        }
        //System.out.println(l + " " + r + " " + u + " " + v);
        //System.out.println(result);
        return result;
    }
    
    public static void main(String[] args) throws Exception
    {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        Scanner sc = new Scanner(System.in);
        String[] tokens = reader.readLine().split(" ");
        n = Long.parseLong(tokens[0]);
        m = Integer.parseInt(tokens[1]);
        mod = Integer.parseInt(tokens[2]);
        
        for (int i = 0; i < m; i++)
        {
            if (System.currentTimeMillis() - start > 5000)
                return;
            tokens = reader.readLine().split(" ");
            long l = Long.parseLong(tokens[0]);
            long r = Long.parseLong(tokens[1]);
            long u = Long.parseLong(tokens[2]);
            long v = Long.parseLong(tokens[3]);
            System.out.println(query(l, r, u, v, 1, n, 1, 1));
        }
    }
}