D - Strongly Connected City

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本文探讨了在给定的城市网格中，通过设置单向街道实现强连通性的问题，即从任意一点出发都能到达城市的任何其他点。通过输入描述街道方向的字符串，判断所给的街道布局是否满足条件。

D - Strongly Connected City

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

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Description

Imagine a city with n horizontal streets crossing m vertical streets, forming an (n - 1) × (m - 1) grid. In order to increase the traffic flow, mayor of the city has decided to make each street one way. This means in each horizontal street, the traffic moves only from west to east or only from east to west. Also, traffic moves only from north to south or only from south to north in each vertical street. It is possible to enter a horizontal street from a vertical street, or vice versa, at their intersection.

$\text{[math]}$

The mayor has received some street direction patterns. Your task is to check whether it is possible to reach any junction from any other junction in the proposed street direction pattern.

Input

The first line of input contains two integers n and m, (2 ≤ n, m ≤ 20), denoting the number of horizontal streets and the number of vertical streets.

The second line contains a string of length n, made of characters '<' and '>', denoting direction of each horizontal street. If the i-th character is equal to '<', the street is directed from east to west otherwise, the street is directed from west to east. Streets are listed in order from north to south.

The third line contains a string of length m, made of characters '^' and 'v', denoting direction of each vertical street. If the i-th character is equal to '^', the street is directed from south to north, otherwise the street is directed from north to south. Streets are listed in order from west to east.

Output

If the given pattern meets the mayor's criteria, print a single line containing "YES", otherwise print a single line containing "NO".

Sample Input

Input

3 3
><>
v^v

Output

NO

Input

4 6
<><>
v^v^v^

Output

YES

小结：

果然，一个人没有脑洞是不行的...脑洞大开后很轻易的就可以发现这道题是一道彻彻底底的水题！！

以下是AC代码：

#include<stdio.h>
#include<string.h>
char x[30],y[30];
int main()
{
    int row,col;
    scanf("%d%d",&row,&col);
    scanf("%s",x+1);
    scanf("%s",y+1);
    if((x[1]=='<'&&y[1]=='^')||(x[row]=='<'&&y[1]=='v')||(x[1]=='>'&&y[col]=='^')||(x[row]=='>'&&y[col]=='v'))
    printf("NO\n");
    else
    printf("YES\n");
    return 0;
}