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

/**
 * Generated by Contest helper plug-in
 * Actual solution is at the bottom
 */
public class Main {
	public static void main(String[] args) {
		InputReader in = new StreamInputReader(System.in);
		PrintWriter out = new PrintWriter(System.out);
		run(in, out);
	}

	public static void run(InputReader in, PrintWriter out) {
		Solver solver = new TaskD();
		solver.solve(1, in, out);
		Exit.exit(in, out);
	}
}

abstract class InputReader {
	private boolean finished = false;

	public abstract int read();

	public int readInt() {
		int c = read();
		while (isSpaceChar(c))
			c = read();
		int sgn = 1;
		if (c == '-') {
			sgn = -1;
			c = read();
		}
		int res = 0;
		do {
			if (c < '0' || c > '9')
				throw new InputMismatchException();
			res *= 10;
			res += c - '0';
			c = read();
		} while (!isSpaceChar(c));
		return res * sgn;
	}

	public long readLong() {
		int c = read();
		while (isSpaceChar(c))
			c = read();
		int sgn = 1;
		if (c == '-') {
			sgn = -1;
			c = read();
		}
		long res = 0;
		do {
			if (c < '0' || c > '9')
				throw new InputMismatchException();
			res *= 10;
			res += c - '0';
			c = read();
		} while (!isSpaceChar(c));
		return res * sgn;
	}

	public String readString() {
		int c = read();
		while (isSpaceChar(c))
			c = read();
		StringBuffer res = new StringBuffer();
		do {
			res.appendCodePoint(c);
			c = read();
		} while (!isSpaceChar(c));
		return res.toString();
	}

	private boolean isSpaceChar(int c) {
		return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
	}

	public void setFinished(boolean finished) {
		this.finished = finished;
	}

	public abstract void close();
}

class StreamInputReader extends InputReader {
	private InputStream stream;
	private byte[] buf = new byte[1024];
	private int curChar, numChars;

	public StreamInputReader(InputStream stream) {
		this.stream = stream;
	}

	public int read() {
		if (numChars == -1)
			throw new InputMismatchException();
		if (curChar >= numChars) {
			curChar = 0;
			try {
				numChars = stream.read(buf);
			} catch (IOException e) {
				throw new InputMismatchException();
			}
			if (numChars <= 0)
				return -1;
		}
		return buf[curChar++];
	}

	public void close() {
		try {
			stream.close();
		} catch (IOException ignored) {
		}
	}
}

class Exit {
	private Exit() {
	}

	public static void exit(InputReader in, PrintWriter out) {
		in.setFinished(true);
		in.close();
		out.close();
	}
}

interface Solver {
	public void solve(int testNumber, InputReader in, PrintWriter out);
}

class TaskD implements Solver {
	private long doubleMod;

	public void solve(int testNumber, InputReader in, PrintWriter out) {
		long size = in.readLong();
		int count = in.readInt();
		long mod = in.readInt();
		doubleMod = 2 * mod;
		for (int i = 0; i < count; i++) {
			long left = in.readLong() - 1;
			long right = in.readLong();
			long min = in.readLong();
			long max = Math.min(in.readLong(), size);
			out.println(query(0, size, 1L, 1L, left, right, min, max, mod));
		}
	}

	private long query(long from, long to, long start, long shift, long left, long right, long min, long max, long mod) {
		if (from >= right || to <= left)
			return 0;
		if (from >= left && to <= right) {
			long plus = Math.max(0, min - start + shift - 1) / shift;
			start += plus * shift;
			if (start > max)
				return 0;
			long count = (max - start) / shift;
			long end = start + count * shift;
			count++;
			long first = (start + end) % doubleMod;
			count %= doubleMod;
			return (first * count) % doubleMod / 2;
		}
		long middle = (from + to + 1) / 2;
		long result = query(from, middle, start, shift << 1, left, right, min, max, mod) +
			query(middle, to, start + shift, shift << 1, left, right, min, max, mod);
		if (result >= mod)
			result -= mod;
		return result;
	}


}