CodeForces

Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them.

Chilly Willy wants to find the minimum number of length n, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that.

A number's length is the number of digits in its decimal representation without leading zeros.

Input

A single input line contains a single integer n (1 ≤ n ≤ 10⁵).

Output

Print a single integer — the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist.

Example

Input

1

Output

-1

Input

5

Output

10080

看到倒是有规律解法，没想到还是有这种暴力解，也是按规律，比赛时没网这个方向想

代码：

#include<iostream>#include<string>#include<cstdio>#include<algorithm>#include<cmath>#include<iomanip>#include<queue>#include<cstring>#include<map>using namespace std;typedef long long ll;#define M 10005int n;int main(){    int i;    scanf("%d",&n);    if(n<=2)    {        printf("-1\n");    }else if(n==3)    {        printf("210\n");    }else if(n==4)    {        printf("1050\n");    }else {        ll ans=50;        for(i=5;i<=n;i++)        {            ans=ans*10;            while(ans-210>=0)            {                ans-=210;            }        }        printf("1");        int len=log10(ans)+1;        for(int i=1;i<=n-len-1;i++)        {            printf("0");        }        printf("%I64d\n",ans);    }    return 0;}