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:
//package round314;
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;

public class F {
	InputStream is;
	PrintWriter out;
	String INPUT = "";
	
	void solve()
	{
		int n = ni(), K = ni();
		int[] cl = new int[K];
		int[] sign = new int[K];
		int[] cr = new int[K];
		for(int i = 0;i < K;i++){
			cl[i] = ni()-1;
			String x = ns();
			if(x.equals("<")){
				sign[i] = -2;
			}else if(x.equals("<=")){
				sign[i] = -1;
			}else if(x.equals("=")){
				sign[i] = 0;
			}else if(x.equals(">=")){
				sign[i] = 1;
			}else if(x.equals(">")){
				sign[i] = 2;
			}else{
				throw new RuntimeException(x);
			}
			cr[i] = ni()-1;
		}
		
		long[][] dp = new long[n+1][2*n+1]; // [phase][#left]
		dp[0][0] = 1;
		for(int i = 0;i < n;i++){
			for(int j = 0;j <= 2*n;j++){
				if(dp[i][j] == 0)continue;
				int L = j;
				int R = 2*n-2*i+j-1;
				inner:
				for(int k = 0;k <= (i < n-1 ? 2 : 0);k++){
					for(int c = 0;c < K;c++){
						int vl = 0;
						if(cl[c] < L || cl[c] > R){
							vl = -1;
						}else if(L + k <= cl[c] && cl[c] <= R - (2-k)){
							vl = 1;
						}
						int vr = 0;
						if(cr[c] < L || cr[c] > R){
							vr = -1;
						}else if(L + k <= cr[c] && cr[c] <= R - (2-k)){
							vr = 1;
						}
						if(vl == vr && vl == 0){
							if(sign[c] == -1 || sign[c] == 0 || sign[c] == 1){
							}else{
								continue inner;
							}
						}else if(vl < vr){
							if(sign[c] < 0){
							}else{
								continue inner;
							}
						}else if(vl > vr){
							if(sign[c] > 0){
							}else{
								continue inner;
							}
						}
					}
					dp[i+1][j+k] += dp[i][j];
				}
			}
		}
		long ret = 0;
		for(int i = 0;i <= 2*n;i++){
			ret += dp[n][i];
		}
		out.println(ret);
	}
	
	void run() throws Exception
	{
		is = oj ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);
		
		long s = System.currentTimeMillis();
		solve();
		out.flush();
		tr(System.currentTimeMillis()-s+"ms");
	}
	
	public static void main(String[] args) throws Exception { new F().run(); }
	
	private byte[] inbuf = new byte[1024];
	private int lenbuf = 0, ptrbuf = 0;
	
	private int readByte()
	{
		if(lenbuf == -1)throw new InputMismatchException();
		if(ptrbuf >= lenbuf){
			ptrbuf = 0;
			try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
			if(lenbuf <= 0)return -1;
		}
		return inbuf[ptrbuf++];
	}
	
	private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
	private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
	
	private double nd() { return Double.parseDouble(ns()); }
	private char nc() { return (char)skip(); }
	
	private String ns()
	{
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}
	
	private char[] ns(int n)
	{
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while(p < n && !(isSpaceChar(b))){
			buf[p++] = (char)b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}
	
	private char[][] nm(int n, int m)
	{
		char[][] map = new char[n][];
		for(int i = 0;i < n;i++)map[i] = ns(m);
		return map;
	}
	
	private int[] na(int n)
	{
		int[] a = new int[n];
		for(int i = 0;i < n;i++)a[i] = ni();
		return a;
	}
	
	private int ni()
	{
		int num = 0, b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private long nl()
	{
		long num = 0;
		int b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private boolean oj = System.getProperty("ONLINE_JUDGE") != null;
	private void tr(Object... o) { if(!oj)System.out.println(Arrays.deepToString(o)); }
}