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 com.sun.org.apache.bcel.internal.generic.NEW;

import java.io.*;
import java.math.BigInteger;
import java.util.*;

public class D
{
    private static int MOD;
    private static BigInteger BIG_MOD;
    public static void main(String[] strings) throws IOException
    {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter writer = new PrintWriter(new OutputStreamWriter(System.out));

        StringTokenizer tokenizer = new StringTokenizer(reader.readLine());
        long n = Long.parseLong(tokenizer.nextToken());
        int q = Integer.parseInt(tokenizer.nextToken());
        MOD = Integer.parseInt(tokenizer.nextToken());
        BIG_MOD = new BigInteger(MOD+"");

        while(q-- > 0)
        {
            tokenizer = new StringTokenizer(reader.readLine());
            long l = Long.parseLong(tokenizer.nextToken()) - 1;
            long r = Long.parseLong(tokenizer.nextToken()) - 1;
            long u = Long.parseLong(tokenizer.nextToken());
            long v = Long.parseLong(tokenizer.nextToken());
            if(u > n)
            {
                writer.println(0);
                continue;
            }
            v = Math.min(v, n);
            u--;
            v--;

            long res = solve(0, 1, 0, n-1, l, r, u, v);
            res += MOD;
            res %= MOD;
            writer.println(res);
        }

        writer.flush();
        writer.close();
    }

    private static long solve(long curr, long depth, long treeLeft, long treeRight, long l, long r, long u, long v)
    {
        l = Math.max(l, treeLeft);
        r = Math.min(r, treeRight);
        if(l > r)
            return 0;
        if(l == treeLeft && r == treeRight)
        {
            long res = calc(curr, depth, v) - calc(curr, depth, u-1);
            res %= MOD;
            return res;
        }

        long treeMid = treeLeft + treeRight >> 1;
        long res = solve(curr, depth<<1, treeLeft, treeMid, l, r, u, v) + solve(curr+depth, depth<<1, treeMid+1, treeRight, l, r, u, v);
        res %= MOD;
        return res;

    }

    private static long calc(long curr, long depth, long bound)
    {
        long tmp = bound;
        tmp -= curr;
        if(tmp < 0)
            return 0;
        tmp /= depth;
        long half = tmp;
        long other = tmp+1;
        if((half & 1) != 0)
        {
            half++;
            other--;
        }
        half >>= 1;

        long res = (((tmp+1)%MOD)*(curr%MOD))%MOD;
        res += (((half%MOD)*(other%MOD)%MOD)*(depth%MOD))%MOD;
        res %= MOD;
        res += tmp+1;
        res %= MOD;

        return res;
    }
}