Problem: King of Berland Berl IV has recently died. Hail Berl V! As a sign of the highest achievements of the deceased king the new king decided to build a mausoleum with Berl IV's body on the main square of the capital.

The mausoleum will be constructed from 2n blocks, each of them has the shape of a cuboid. Each block has the bottom base of a 1 × 1 meter square. Among the blocks, exactly two of them have the height of one meter, exactly two have the height of two meters, ..., exactly two have the height of n meters.

The blocks are arranged in a row without spacing one after the other. Of course, not every arrangement of blocks has the form of a mausoleum. In order to make the given arrangement in the form of the mausoleum, it is necessary that when you pass along the mausoleum, from one end to the other, the heights of the blocks first were non-decreasing (i.e., increasing or remained the same), and then — non-increasing (decrease or remained unchanged). It is possible that any of these two areas will be omitted. For example, the following sequences of block height meet this requirement:

  * [1, 2, 2, 3, 4, 4, 3, 1]; 
  * [1, 1]; 
  * [2, 2, 1, 1]; 
  * [1, 2, 3, 3, 2, 1]. 



Suddenly, k more requirements appeared. Each of the requirements has the form: "h[xi] signi h[yi]", where h[t] is the height of the t-th block, and a signi is one of the five possible signs: '=' (equals), '<' (less than), '>' (more than), '<=' (less than or equals), '>=' (more than or equals). Thus, each of the k additional requirements is given by a pair of indexes xi, yi (1 ≤ xi, yi ≤ 2n) and sign signi.

Find the number of possible ways to rearrange the blocks so that both the requirement about the shape of the mausoleum (see paragraph 3) and the k additional requirements were met.

Input

The first line of the input contains integers n and k (1 ≤ n ≤ 35, 0 ≤ k ≤ 100) — the number of pairs of blocks and the number of additional requirements.

Next k lines contain listed additional requirements, one per line in the format "xi signi yi" (1 ≤ xi, yi ≤ 2n), and the sign is on of the list of the five possible signs.

Output

Print the sought number of ways.

Examples

Input

3 0


Output

9


Input

3 1
2 &gt; 3


Output

1


Input

4 1
3 = 6


Output

3

Solution (in JAVA):

Solution:
import java.util.*;
import java.util.Map.Entry;
import java.io.*;
import java.lang.*;
import java.math.*;

import static java.lang.Math.*;

public class Solution implements Runnable {
	
	long[][] dp;
	int[][] a; // > 1 < -1 = 0 >= 2 <= -2
	int n;
	
	boolean check(int i, int j, int lf, int rg) {
		if (a[i][j] == 1 || a[i][j] == -1)
			return false;
		for (int k = 0; k < 2 * n; k++) {
			if (k != i && k != j) {
				if (k < lf || k > rg) {
					if (a[i][k] == -1 || a[i][k] == 0 || a[i][k] == -2) {
						return false;
					}
					if (a[j][k] == -1 || a[j][k] == 0 || a[j][k] == -2) {
						return false;
					}
				} else {
					if (a[i][k] == 1 || a[i][k] == 0 || a[i][k] == 2) {
						return false;
					}
					if (a[j][k] == 1 || a[j][k] == 0 || a[j][k] == 2) {
						return false;
					}
				}
			}
		}
		return true;
	}
	
	long go(int lf, int rg) {
		if (dp[lf][rg] != -1)
			return dp[lf][rg];
		if (rg - lf + 1 == 2) {
			if (a[lf][rg] == -3 || (a[lf][rg] % 2) == 0)
				dp[lf][rg] = 1;
			else
				dp[lf][rg] = 0;
			return dp[lf][rg];
		}
		dp[lf][rg] = 0;
		if (check(lf, lf + 1, lf + 2, rg)) {
			dp[lf][rg] += go(lf + 2, rg);
		}
		if (check(lf, rg, lf + 1, rg - 1)) {
			dp[lf][rg] += go(lf + 1, rg - 1);
		}
		if (check(rg - 1, rg, lf, rg - 2)) {
			dp[lf][rg] += go(lf, rg - 2);
		}
		return dp[lf][rg];
	}
	
    void solve() throws Exception {
    	n = sc.nextInt();
    	int k = sc.nextInt();
    	dp = new long[2 * n][2 * n];
    	a = new int[2 * n][2 * n];
    	for (int i = 0; i < 2 * n; i++) {
    		Arrays.fill(a[i], -3);
    		Arrays.fill(dp[i], -1);
    	}
    	while (k-- > 0) {
    		int i = sc.nextInt() - 1;
    		String s = sc.nextToken();
    		int j = sc.nextInt() - 1;
    		if (s.equals(">")) {
    			a[i][j] = 1;
    			a[j][i] = -1;
    		}
    		if (s.equals("<")) {
    			a[i][j] = -1;
    			a[j][i] = 1;
    		}
    		if (s.equals("=")) {
    			a[i][j] = a[j][i] = 0;
    		}
    		if (s.equals(">=")) {
    			a[i][j] = 2;
    			a[j][i] = -2;
    		}
    		if (s.equals("<=")) {
    			a[i][j] = -2;
    			a[j][i] = 2;
    		}
    	}
    	out.println(go(0, 2 * n - 1));
    }
    
    BufferedReader in;
    PrintWriter out;
    FastScanner sc;
    
    final String INPUT_FILE = "stdin";
    final String OUTPUT_FILE = "stdout";
    
    static Throwable throwable;
    
    public static void main(String[] args) throws Throwable {
        Thread thread = new Thread(null, new Solution(), "", (1 << 26));
        thread.start();
        thread.join();
        thread.run();
        if (throwable != null)
            throw throwable;
    }
    
    
    public void run() {
        try {
            if (INPUT_FILE.equals("stdin"))
                in = new BufferedReader(new InputStreamReader(System.in));
            else
                in = new BufferedReader(new FileReader(INPUT_FILE));
            if (OUTPUT_FILE.equals("stdout"))
                out = new PrintWriter(System.out);
            else
                out = new PrintWriter(new FileWriter(OUTPUT_FILE));
            sc = new FastScanner(in);
            solve();
        } catch (Exception e) {
            throwable = e;
        } finally {
            out.close();
        }
    }
}

class FastScanner {
    BufferedReader reader;
    StringTokenizer strTok;
    
    FastScanner(BufferedReader reader) {
        this.reader = reader;
    }
    
    public String nextToken() throws Exception {
        while (strTok == null || !strTok.hasMoreTokens())
            strTok = new StringTokenizer(reader.readLine());
        return strTok.nextToken();
    }
    
    public int nextInt() throws Exception {
        return Integer.parseInt(nextToken());
    }
    
    public long nextLong() throws Exception {
        return Long.parseLong(nextToken());
    }
    
    public double nextDouble() throws Exception {
        return Double.parseDouble(nextToken());
    }
}