CodeForces 248B Chilly Willy

B. Chilly Willy

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

input

1

output

-1

input

5

output

10080

规律题

#include <cstdio>
#include <algorithm>
using namespace std;

const int maxn = 100000 + 10;

int main()
{
    int n;
    int ans;
    while (scanf("%d", &n) != EOF){
        if (n < 3)
            printf("-1\n");
        else if (n == 3)
            printf("210\n");
        else{
            printf("1");
            for (int i = 0; i < n - 4; i++){
                printf("0");
            }
            int ans = (n - 4) % 6;
            if (ans == 0)
                printf("05");
            if (ans == 1)
                printf("08");
            if (ans == 2)
                printf("17");
            if (ans == 3)
                printf("02");
            if (ans == 4)
                printf("20");
            if (ans == 5)
                printf("11");
            printf("0\n");
        }
    }
    return 0;
}

#include<cstdio>
int i,p,n;
int main(){
	scanf("%d",&n);
	if(n<=3)printf("%d\n",n<=2?-1:210);
	else{
		printf("1");
		for(p=50,i=1;i<=n-4;++i)
			printf("0"),p=p*10%210;
		printf("%03d\n",p);
	}
}