While swimming at the beach, Mike has accidentally dropped his cellphone into the water. There was n

最新推荐文章于 2019-02-12 01:30:22 发布

原创最新推荐文章于 2019-02-12 01:30:22 发布 · 658 阅读

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本文探讨了通过手指移动轨迹来识别手机键盘上输入的电话号码的问题。具体来说，考虑到只有十个数字按键的老式键盘布局，分析了如何判断两次输入是否对应相同的按键移动路径。

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:

$\text{[math]}$

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":

$\text{[math]}$ $\text{[math]}$

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.

Sample Input

Input

3
586

Output

NO

Input

2
09

Output

NO

Input

9
123456789

Output

YES

Input

3
911

Output

YES

#include<stdio.h>
#include<iostream>
#include<cstring>
using namespace std;


int main()
{
    int n, i;
    char s[10000];
    scanf("%d", &n);
    /*for(i=0; i<n; i++)
        scanf("%c", &s[i]);*/
    scanf("%s", s);
    int shang=0, xia=0, zuo=0, you=0;
    for(i=0; i<n; i++)
    {
        if(s[i] == '0')
        {
            xia = zuo = you = 1;
        }
        if(s[i] == '1' || s[i] == '2' || s[i] == '3')
        {
            shang = 1;
        }
        if(s[i] == '1' || s[i] == '4' || s[i] == '7')
        {
            zuo = 1;
        }
        if(s[i] == '3' || s[i] == '6' || s[i] == '9')
        {
            you = 1;
        }
        if(s[i] == '7' || s[i] == '9')
        {
            xia = 1;
        }
    }
    if(shang==1 && xia==1 && zuo==1 && you==1)
        printf("YES\n");
    else
        printf("NO\n");
    return 0;
}