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:
/*
Alfonso2 Peterssen(mukel)
Codeforces Beta Round #87 (Div. 1 Only)
*/
import java.util.*;
import java.io.*;
import java.math.*;

public class D implements Runnable
{
	long N, Q, MOD, begin, end, lb, ub;

	long sum(long f ,long gap, long n)
	{
		++n;

		f %= MOD;
		n %= MOD;
		gap %= MOD;
		
		return (n * f % MOD + n*(n - 1)/2 % MOD * gap) % MOD;
	}

	long getSum(long lo, long hi, long roof, long first, long gap)
	{
		if (lo > hi || lo > end || hi < begin || first > roof)
			return 0;		

		if (lo >= begin && hi <= end)
		{
			long n = (roof - first) / gap;
			n = Math.min(n, hi - lo);
			return sum(first, gap, n);
		}

		long mid = (lo + hi) / 2;

		return
			(getSum(lo, mid, roof, first, gap * 2) +
			getSum(mid + 1, hi, roof, first + gap, gap * 2)) % MOD;
	}
	
	void solve() throws IOException
	{
		N = nextLong();
		Q = nextLong();
		MOD = nextLong();

		while (Q-- > 0)
		{
			begin = nextLong();
			end = nextLong();
			lb = nextLong();
			ub = nextLong();

			writer.println((getSum(1, N, ub, 1, 1) - getSum(1, N, lb - 1, 1, 1) + MOD) % MOD);
		}		
	}

	public static void main(String args[])
	{
		new D().run();
	}

	BufferedReader reader;
	PrintWriter writer;
	StringTokenizer tokenizer;

	public void run()
	{
		try
		{
			reader = new BufferedReader(new InputStreamReader(System.in));
			tokenizer = null;
			writer = new PrintWriter(System.out);
			solve();
			reader.close();
			writer.close();			
		}
		catch (Exception e)
		{
			e.printStackTrace();
			System.exit(1);
		}		
	}

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

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

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

	BigInteger nextBigInteger() throws IOException
	{
		return new BigInteger(nextToken());
	}

	String nextToken() throws IOException
	{
		while (tokenizer == null || !tokenizer.hasMoreTokens())
		{
			tokenizer = new StringTokenizer(reader.readLine());
		}
		return tokenizer.nextToken();
	}
}