Problem: Let us remind you the rules of a very popular game called "Snake" (or sometimes "Boa", "Python" or "Worm").

The game field is represented by an n × m rectangular table. Some squares of the field are considered impassable (walls), all other squares of the fields are passable.

You control a snake, the snake consists of segments. Each segment takes up exactly one passable square of the field, but any passable square contains at most one segment. All segments are indexed by integers from 1 to k, where k is the snake's length. The 1-th segment is the head and the k-th segment is the tail. For any i (1 ≤ i < k), segments with indexes i and i + 1 are located in the adjacent squares of the field, that is, these squares share a common side.

One of the passable field squares contains an apple. The snake's aim is to reach the apple and eat it (that is, to position its head in the square with the apple).

The snake moves throughout the game. During one move the snake can move its head to an adjacent field square. All other segments follow the head. That is, each segment number i (1 < i ≤ k) moves to the square that has just had segment number i - 1. Consider that all segments including the head move simultaneously (see the second test sample). If the snake's head moves to an unpassable square or to the square, occupied by its other segment, the snake dies. That's why we will consider such moves unvalid.

Your task is to determine the minimum number of valid moves that the snake needs to reach the apple.

Input

The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 15) — the number of rows and columns of the game field.

Next n lines describe the game field. Each of these lines contains m characters. Character "#" represents a wall, "." is a passable square, "@" is an apple. The snake's first segment is represented by character "1", the second one segment — by character "2" and so on.

The game field description doesn't contain any characters besides "#', ".", "@" and digits (except 0). It is guaranteed that the described field is correct. It is guaranteed that the described field contains exactly one apple and exactly one snake, the snake's length is at least 3 and at most 9.

Output

Print a single integer to the output — the minimum number of moves needed to reach the apple. If the snake can't reach the apple, print -1.

Examples

Input

4 5
##...
..1#@
432#.
...#.


Output

4


Input

4 4
#78#
.612
.543
..@.


Output

6


Input

3 2
3@
2#
1#


Output

-1

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.StringTokenizer;


public class CodeD 
{
	static class Scanner
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		
		String nextLine()
		{
			try { return br.readLine(); } catch(Exception e) { throw new RuntimeException(e); }
		}
		
		String next()
		{
			while(!st.hasMoreTokens())
			{
				String line = nextLine();
				if(line == null) return null;
				st = new StringTokenizer(line);
			}
			return st.nextToken();
		}
		
		int nextInt()
		{
			return Integer.parseInt(next());
		}
		
		long nextLong()
		{
			return Long.parseLong(next());
		}
		
		
		private double nextDouble() 
		{
			return Double.parseDouble(next());
		}

		int[] nextIntArray(int n)
		{
			int[] nA = new int[n];
			for(int i = 0; i < n; i++)
				nA[i] = nextInt();
			return nA;
		}
		
		Integer[] nextIntegerArray(int n)
		{
			Integer[] nA = new Integer[n];
			for(int i = 0; i < n; i++)
				nA[i] = nextInt();
			return nA;
		}
		
		long[] nextLongArray(int n)
		{
			long[] nA = new long[n];
			for(int i = 0; i < n; i++)
				nA[i] = nextLong();
			return nA;
		}
		
		Long[] nextLongOArray(int n)
		{
			Long[] nA = new Long[n];
			for(int i = 0; i < n; i++)
				nA[i] = nextLong();
			return nA;
		}
		
		double[] nextDoubleArray(int n)
		{
			double[] nA = new double[n];
			for(int i = 0; i < n; i++)
				nA[i] = nextDouble();
			return nA;
		}

		Double[] nextDoubleOArray(int n)
		{
			Double[] nA = new Double[n];
			for(int i = 0; i < n; i++)
				nA[i] = nextDouble();
			return nA;
		}
		
		String[] nextStringArray(int n)
		{
			String[] nA = new String[n];
			for(int i = 0; i < n; i++)
				nA[i] = next();
			return nA;
		}
	}
	
	static int nSegmentos;
	
	static class Culebra
	{
		byte[] segmentos;
		boolean termino;
		int cuenta;
		
		public Culebra(byte[] is)
		{
			segmentos = is;
		}
		
		static final int[][] diffs = new int[][] {{1, 0}, {-1, 0}, {0, -1}, {0, 1}};
		
		Culebra generarHijos(LinkedList <Culebra> c, HashSet <Culebra> h)
		{
			for(int[] d : diffs)
			{
				int iCab = segmentos[0] + d[0];
				int jCab = segmentos[1] + d[1];
				if(valido(iCab, jCab))
				{
					Culebra hijo = generarHijo(iCab, jCab);
					if(!h.contains(hijo))
					{
						h.add(hijo);
						c.add(hijo);
						if(hijo.termino)
							return hijo;
					}
				}
			}
			return null;
		}

		private Culebra generarHijo(int iCab, int jCab) 
		{
			byte[] sNuevos = segmentos.clone();
			Culebra nueva = new Culebra(sNuevos);
			sNuevos[0] = (byte) iCab;
			sNuevos[1] = (byte) jCab;
			for(int i = 1; i < nSegmentos; i++)
			{
				sNuevos[i << 1] = segmentos[(i - 1) << 1];
				sNuevos[(i << 1) + 1] = segmentos[((i - 1) << 1) + 1];
			}
			nueva.cuenta = cuenta + 1;
			nueva.termino = original[iCab][jCab] == '@';
			return nueva;
		}

		static final boolean[][] tieneSegmentos = new boolean[15][15];
		
		private boolean valido(int iCab, int jCab)
		{
			boolean bien = iCab >= 0 && iCab < original.length && jCab >= 0 && jCab < original[0].length && original[iCab][jCab] != '#';
			if(!bien)
				return bien;
			for(int i = 0; i < nSegmentos - 1; i++)
				tieneSegmentos[segmentos[i << 1]][segmentos[(i << 1) + 1]] = true;
			if(tieneSegmentos[iCab][jCab])
				bien = false;
			for(int i = 0; i < nSegmentos - 1; i++)
				tieneSegmentos[segmentos[i << 1]][segmentos[(i << 1) + 1]] = false;
			return bien;
		}

		@Override
		public int hashCode()
		{
			return Arrays.hashCode(segmentos);
		}

		@Override
		public boolean equals(Object obj)
		{
			return Arrays.equals(segmentos, ((Culebra) obj).segmentos);
		}
		
//		@Override
//		public String toString() 
//		{
//			for(int i = 0; i < nSegmentos; i++)
//				original[segmentos[i << 1]][segmentos[(i << 1) + 1]] = (char) ('1' + i);
//			String s = "";
//			for(char[] c : original)
//				s += new String(c) + "\n";
//			for(int i = 0; i < nSegmentos; i++)
//				original[segmentos[i << 1]][segmentos[(i << 1) + 1]] = '.';
//			return s;
//		}
	}
	
	static byte[] calcularSegmentos(char[][] tablero)
	{
		byte[] segmentos = new byte[nSegmentos << 1];
		for(int i = 0; i < tablero.length; i++)
		{
			for(int j = 0; j < tablero[0].length; j++)
				if(tablero[i][j] >= '1' && tablero[i][j] <= '9')
				{
					int indice = (tablero[i][j] - '1') << 1;
					segmentos[indice] = (byte) i;
					segmentos[indice + 1] = (byte) j;
					tablero[i][j] = '.';
				}
		}
		return segmentos;
	}
	
	static char[][] original;
	
	public static void main(String[] args)
	{
		Scanner sc = new Scanner();
		int n = sc.nextInt();
		int m = sc.nextInt();
		original = new char[n][];
		for(int i = 0; i < n; i++)
			original[i] = sc.next().toCharArray();
		for(int i = 0; i < n; i++)
			for(int j = 0; j < m; j++)
				if(original[i][j] >= '1' && original[i][j] <= '9')
					nSegmentos = Math.max(nSegmentos, original[i][j] - '0');
		Culebra inicial = new Culebra(calcularSegmentos(original));
		inicial.cuenta = 0;
		LinkedList <Culebra> c = new LinkedList <Culebra> ();
		HashSet <Culebra> h = new HashSet <Culebra> ();
		c.add(inicial);
		h.add(inicial);
		while(!c.isEmpty())
		{
			Culebra actual = c.poll();
			Culebra posible = actual.generarHijos(c, h);
			if(posible != null)
			{
				System.out.println(posible.cuenta);
				return;
			}
		}
		System.out.println("-1");
	}
}