codeforces 2184 B. Chilly Willy（规律）

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.

Sample test(s)

Input

1

Output

-1

Input

5

Output

10080

 

题目大意，求出能够把2,3,5,7整除的数，前提是输出该数的位数，并且是最小的。

比赛的时候，也感觉到了是规律题，但是没找到- -！我找的是数字为210的倍数，但是并没有规律，这题是找输入N的规律。

 

 

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<map>
#include<queue>
#include<cmath>
#define LL long long
#define inf 0x3f3f3f3f
LL a[10000010];
LL b[100010];
int fla[1000010];
using namespace std;
int main()
{
    LL n,i,j;
    while(~scanf("%lld",&n))
    {
        if(n<=2)
        {
            printf("-1\n");
            continue;
        }
        else if(n==3)
        {
            printf("210\n");
            continue;
        }
       LL k=n%6;
        printf("1");
        if(k==0)
        {
            for(i=0;i<n-4;i++)
            {
                printf("0");
            }
            printf("170\n");
        }
        else if(k==1)
        {
            for(i=0;i<n-3;i++)
                printf("0");
            printf("20\n");
        }
        else if(k==2)
        {
            for(i=0;i<n-4;i++)
                printf("0");
            printf("200\n");
        }
        else if(k==3)
        {
            for(i=0;i<n-4;i++)
                printf("0");
            printf("110\n");
        }
        else if(k==4)
        {
            for(i=0;i<n-3;i++)
                printf("0");
            printf("50\n");
        }
        else if(k==5)
        {
            for(i=0;i<n-3;i++)
                printf("0");
            printf("80\n");
        }
    }
    return 0;
}