Problem: Ksenia has a chessboard of size n × m. Each cell of the chessboard contains one of the characters: "<", ">", "^", "v", "#". The cells that contain character "#" are blocked. We know that all chessboard cells that touch the border are blocked.

Ksenia is playing with two pawns on this chessboard. Initially, she puts the pawns on the chessboard. One cell of the chessboard can contain two pawns if and only if the cell is blocked. In other cases two pawns can not stand in one cell. The game begins when Ksenia put pawns on the board. In one move, Ksenia moves each pawn to a side adjacent cell in the direction of arrows painted on the cell on which the corresponding pawn sits (if the pawn sits on "#", it does not move). Assume that Ksenia moves pawns simultaneously (see the second test case). 

Of course, Ksenia plays for points. How can one calculate the points per game? Very simply! Let's count how many movements the first pawn made and how many movements the second pawn made, sum these two numbers — it will be the resulting score of the game. 

Ksenia wonders: what is the maximum number of points she can earn (for that, she should place the pawns optimally well early in the game). Help her and find that number. 

Input

The first line contains two integers, n and m (1 ≤ n, m ≤ 2000) — the sizes of the board. Each of the following n lines contains m characters – the board's description. Each character is one of the characters: "<", ">", "^", "v", "#".

It is guaranteed that the border cells of the table are blocked cells (with character "#").

Output

If Ksenia can get infinitely many points, print -1. Otherwise, print the maximum number of points she can get.

Examples

Input

1 1
#


Output

0


Input

3 4
####
#&gt;^{}#
####


Output

3

Input

3 4
####
#&gt;&lt;#
####


Output

-1

Input

7 5
#####
##v##
##v##
#####
##^{}##
##^{}##
#####


Output

4

Input

7 5
#####
##v##
##v##
##&lt;##
##^{}##
##^{}##
#####

Output

5

Solution (in JAVA):

Solution:
import java.io.IOException;
import java.util.Arrays;
import java.io.OutputStream;
import java.util.ArrayDeque;
import java.io.PrintWriter;
import java.util.Queue;
import java.util.Collection;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author Mahmoud Aladdin <aladdin3>
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskD solver = new TaskD();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskD {

    int[] dx = new int[]{-1, 0, 1, 0};
    int[] dy = new int[]{0, -1, 0, 1};
    int[] check = new int[]{2, 3, 0, 1};
    int[][] grid;
    int max1 = 0;
    int max2 = 0;
    int start1 = 0;

    public void solve(int testNumber, InputReader jin, PrintWriter jout) {
        int m = jin.int32();
        int n = jin.int32();
        int[][] val = new int[m][n];
        int[][] start = new int[m][n];
        for(int[] r: val) Arrays.fill(r, -1);
        for(int[] r: start) Arrays.fill(r, -1);
        grid = new int[m][n];
        Queue<Integer> que = new ArrayDeque<Integer>();
        for(int i = 0; i < m; i++) {
            String token = jin.token();
            for(int j = 0; j < n; j++) {
                switch(token.charAt(j)) {
                    case '#': grid[i][j] = -1; break;
                    case '^': grid[i][j] = 0; break;
                    case '<': grid[i][j] = 1; break;
                    case 'v': grid[i][j] = 2; break;
                    case '>': grid[i][j] = 3; break;
                }
                if(grid[i][j] == -1) {
                    que.add(i * n + j);
                    val[i][j] = 0;
                }
            }
        }

        while(!que.isEmpty()) {
            int p = que.poll();
            int x = p / n;
            int y = p % n;
            int cnt = 0;
            for(int i = 0; i < 4; i++) {
                int nx = x + dx[i];
                int ny = y + dy[i];
                if(validate(i, nx, ny, m, n)) {
                    cnt++;
                    que.add(nx * n + ny);
                    val[nx][ny] = val[x][y] + 1;
                    start[nx][ny] = (start[x][y] == -1)? (nx * n + ny) : start[x][y];
                }
            }
            if(grid[x][y] != -1) {
                if(val[x][y] > max1) {
                   max2 = max1;
                   max1 = val[x][y];
                   start1 = start[x][y];
                } else {
                    if(val[x][y] == max1 && start1 == start[x][y])
                        continue;
                    if(val[x][y] > max2)
                        max2 = val[x][y];
                }
            }
        }

        boolean inf = false;
        all: for(int i = 0; i < m; i++) {
            for(int j = 0; j < n; j++) {
                if(val[i][j] == -1) {
                    inf = true;
                    break all;
                }
            }
        }

        if(inf) {
            jout.println(-1);
        } else {
            if(max2 == 0 && max1 > 0) max2 = max1 - 1;
            jout.println(max1 + max2);
        }
    }

    private boolean validate(int i, int nx, int ny, int m, int n) {
        if(nx < 0 || nx >= m || ny < 0 || ny >= n)
            return false;
        return check[i] == grid[nx][ny];
    }

}

class InputReader {
    private static final int bufferMaxLength = 1024;
    private InputStream in;
    private byte[] buffer;
    private int currentBufferSize;
    private int currentBufferTop;
    private static final String tokenizers = " \t\r\f\n";
    
    public InputReader(InputStream stream) {
        this.in = stream;
        buffer = new byte[bufferMaxLength];
        currentBufferSize = 0;
        currentBufferTop = 0;
    }
    
    private boolean refill() {
        try {
            this.currentBufferSize = this.in.read(this.buffer);
            this.currentBufferTop = 0;
        } catch(Exception e) {}
        return this.currentBufferSize > 0;
    }
    
    
    private Byte readChar() {
        if(currentBufferTop < currentBufferSize) {
            return this.buffer[this.currentBufferTop++];
        } else {
            if(!this.refill()) {
                return null;
            } else {
                return readChar();
            }
        }
    }

    public String token()  {
        StringBuffer tok = new StringBuffer();
        Byte first;
        while((first = readChar()) != null && (tokenizers.indexOf((char) first.byteValue()) != -1));
        if(first == null) return null;
        tok.append((char)first.byteValue());
        while((first = readChar()) != null && (tokenizers.indexOf((char) first.byteValue()) == -1)) {
            tok.append((char)first.byteValue());
        }
        return tok.toString();
    }
    
    public Integer int32() throws NumberFormatException {
        String tok = token();
        return tok == null? null : Integer.parseInt(tok);
    }

    }