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

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.

直接暴力模拟。

记录一下状态，然后暴力枚举每一个位置，看看有几个位置满足那个状态，如果答案数大于1  那么他就有可能播错电话。..

ccf上有一个类似的..我是不是在做广告...

/* ***********************************************
Author        :guanjun
Created Time  :2016/7/7 22:40:13
File Name     :cf361.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << ;
const double eps=1e-;
using namespace std;int a[maxn];
int b[][]={
    ,,,
    ,,,
    ,,,
    INF,,INF
};
map<int,pair<int,int> >mp;
int main()
{
    int n;
    string s;
    cin>>n;
    cin>>s;
    for(int i=;i<n;i++){
        a[i+]=s[i]-'';
    }
    for(int i=;i<;i++){
        for(int j=;j<;j++){
            mp[b[i][j]]={i,j};
        }
    }
    vector<pair<int,int> >v;
    int t=;
    for(int i=;i<=n;i++){
        int x=mp[a[i]].first-mp[a[t]].first;
        int y=mp[a[i]].second-mp[a[t]].second;
        v.push_back({x,y});
        t=i;
    }
    int ans=;
    int m=v.size();for(int i=;i<;i++){
        for(int j=;j<;j++){
            if(b[i][j]==INF)continue;
            int k;
            int ox=i;
            int oy=j;
            for(k=;k<m;k++){
                int x=ox+v[k].first;
                int y=oy+v[k].second;
                if(x>=||y>=||x<||y<)break;
                if(b[x][y]==INF)break;
                ox=x;
                oy=y;
            }
            if(k==m)ans++;

        }
    }
    if(ans>)puts("NO");
    else puts("YES");
    return ;
}