CQUOJ 9766 Chilly Willy

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.

 

Sample Input

Input

1

Output

-1

Input

5

Output

10080

 

/*
2016年4月24日16:55:26
题意: 给定n  找到十进制数 长度为n的 能同时整除2 3 5 7的最小的数 

要找到210的倍数 很明显只用考虑后三位  最后一位一定为0，
因此找规律可以得到6个数一个循环
"05", "08", "17", "02", "20", "11" 

明显个鬼啊 
*/



# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
# include <queue>
# include <vector>
# include <cmath>
# define LL long long 
# define INF 0x3f3f3f3f
using namespace std;
const int N = 1e5 + 5;

char s[6][3] = {"05", "08", "17", "02", "20", "11"};

int main(void)
{
    int n, tmp, i;
    while (~scanf("%d", &n)){
        if (n <= 2)
            printf("-1\n");
        else if (n == 3)
            printf("210\n");
        else {
            printf("1");
            for (i = 2; i <= n-3; i++)
                printf("0");
            tmp = (n - 4) % 6;
            printf("%s0\n", s[tmp]);
        }
    }
    
    return 0;    
}