Codeforces Round #361 (Div. 2) -- A. Mike and Cellphone （思路题目）

A. Mike and Cellphone

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

While swimming at the beach, Mike has accidentally dropped his cellphone into the water. There was no worry as he bought a cheap replacement phone with an old-fashioned keyboard. The keyboard has only ten digital equal-sized keys, located in the following way:

Together with his old phone, he lost all his contacts and now he can only remember the way his fingers moved when he put some number in. One can formally considerfinger movements as a sequence of vectors connecting centers of keys pressed consecutively to put in a number. For example, the finger movements for number "586" are the same as finger movements for number "253":

Mike has already put in a number by his "finger memory" and started calling it, so he is now worrying, can he be sure that he is calling the correct number? In other words, is there any other number, that has the same finger movements?

Input

The first line of the input contains the only integer n (1 ≤ n ≤ 9) — the number of digits in the phone number that Mike put in.

The second line contains the string consisting of n digits (characters from '0' to '9') representing the number that Mike put in.

Output

If there is no other phone number with the same finger movements and Mike can be sure he is calling the correct number, print "YES" (without quotes) in the only line.

Otherwise print "NO" (without quotes) in the first line.

Examples

Input

3586

Output

NO

Input

209

Output

NO

Input

9123456789

Output

YES

Input

3911

Output

YES

Note

You can find the picture clarifying the first sample case in the statement above.

大体题意:

图形类似于图形密码，输入一个图形密码，问是否存在另一个图形密码，路径与之完全相同！

思路：

思路题目，比赛期间写了很久也没写出来就睡觉了~~  第二天才发现 A的人好少~  还需多练习呀~

借鉴的别人的代码：

设置四个变量  初始化为1 ，表示左边 右边 上边 下边是否可以挪动！

当选择了1 2 3 时，上面不可以走 up = 0

当选择了 3 6 9 右面不可以走 right = 0

当选择了1 4 7 时  左边不可以走 left = 0

当选择了7  9 时  下面不可以走  down  = 0

当选择了0   下面  左边 右边 都不可以走  down  = left = right = 0;

最后看看只要有一个方向可以走，就是NO  否则是YES！

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn = 100 + 10;char s[maxn];int main(){    int n;    while(scanf("%d",&n) == 1){        scanf("%s",s);        int left = 1,right = 1,up = 1,down = 1;        for (int i = 0; i < n; ++i){            char ch = s[i];            if (ch == '1' || ch == '2' || ch == '3')up = 0;            if (ch == '3' || ch == '6' || ch == '9')right = 0;            if (ch == '1' || ch == '4' || ch == '7')left = 0;            if (ch == '7' || ch == '9')down = 0;            if (ch == '0')left = right = down = 0;        }        if (left || right || down || up)printf("NO\n");        else printf("YES\n");    }    return 0;}