Codeforces Round #361 (Div. 2) A. Mike and Cellphone

原题链接

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 consider finger 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

3
586

output

NO

input

2
09

output

NO

input

9
123456789

output

YES

input

3
911

output

YES

Note

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

记录在每个数字的位置能够移动的方向。然后给定输入，判断是否所有的数字都能向同一个方向移动。

#include <bits/stdc++.h>

using namespace std;

int k[10][4] = {
{0, 1, 0, 0},
{0, 0, 1, 1},
{1, 0, 1, 1},
{1, 0, 0, 1},
{0, 1, 1, 1},
{1, 1, 1, 1},
{1, 1, 0, 1},
{0, 1, 1, 0},
{1, 1, 1, 1},
{1, 1, 0, 0}
};
char s[20];
int main(){

int n;

scanf("%d%s", &n, s);
for(int j = 0; j < 4; j++){
int i;
for(i = 0; i < n; i++){
int d = s[i] - '0';
if(k[d][j] == 0)
break;
}
if(i == n){
puts("NO");
return 0;
}
}
puts("YES");
return 0;
}