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 static java.lang.Math.min;
import static java.util.Arrays.deepToString;

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

public class D {

	static long[] get(long n, long MOD, long R, long V) {
		if (R < 0 || V < 0)
			return new long[] { 0, 0 };
		long even = (n + 1) / 2;
		if (R < even - 1) {
			long[] res = get(even, MOD, R, V / 2);
			return new long[] { res[0], res[1] * 2 % MOD };
		} else {
			long cnt = min(even, V / 2 + 1) % MOD;
			long evenRes = ((cnt - 1) * cnt) % MOD;
			long nV = V % 2 == 0 ? V / 2 - 1 : V / 2;
			long[] res = get(n - even, MOD, R - even, nV);
			return new long[] { res[0] + cnt,
					(evenRes + res[0] + 2 * res[1]) % MOD };
		}
	}

	static long sum(long n, long MOD, long R, long V) {
		long[] res = get(n, MOD, R, V);
		return res[0] + res[1];
	}

	static long getRes(long n, long MOD, long L, long R, long U, long V) {
		long ret = (sum(n, MOD, R, V) - sum(n, MOD, L - 1, V))
				- (sum(n, MOD, R, U - 1) - sum(n, MOD, L - 1, U - 1));
		return ((ret % MOD) + MOD) % MOD;
	}

	static long[] solve(long n, int m, long MOD, long[] L, long[] R, long[] U,
			long[] V) throws Exception {
		long[] res = new long[m];
		for (int i = 0; i < m; i++) {
			res[i] = getRes(n, MOD, L[i], R[i], U[i], V[i]);
		}
		return res;
	}

	public static void main(String[] args) {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);

			long n = nextLong();
			int m = nextInt();
			int MOD = nextInt();
			long[] L = new long[m];
			long[] R = new long[m];
			long[] U = new long[m];
			long[] V = new long[m];
			for (int i = 0; i < m; i++) {
				L[i] = nextLong() - 1;
				R[i] = nextLong() - 1;
				U[i] = nextLong() - 1;
				V[i] = nextLong() - 1;
			}

			long[] res = solve(n, m, MOD, L, R, U, V);

			for (long v : res) {
				out.println(v);
			}

			out.close();
		} catch (Throwable e) {
			e.printStackTrace();
			System.exit(1);
		}
	}

	static BufferedReader in;
	static PrintWriter out;
	static StringTokenizer tok;
	static long launchTimer;

	static void debug(Object... o) {
		System.err.println(deepToString(o));
	}

	static void setTime() {
		launchTimer = System.currentTimeMillis();
	}

	static void printTime() {
		System.err.println(System.currentTimeMillis() - launchTimer);
	}

	static void printMemory() {
		System.err.println((Runtime.getRuntime().totalMemory() - Runtime
				.getRuntime().freeMemory()) / 1000 + "kb");
	}

	static boolean hasMoreTokens() throws IOException {
		while (tok == null || !tok.hasMoreTokens()) {
			String line = in.readLine();
			if (line == null) {
				return false;
			}
			tok = new StringTokenizer(line);
		}
		return true;
	}

	static String next() throws IOException {
		return hasMoreTokens() ? tok.nextToken() : null;
	}

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

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

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

	static BigInteger nextBig() throws IOException {
		return new BigInteger(next());
	}
}