CodeForces - 584A Olesya and Rodion （模拟）水

64bit IO Format: %I64d & %I64u

CodeForces - 584A

Olesya and Rodion

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Description

Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.

Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print  - 1.

Input

The single line contains two numbers, n and t (1 ≤ n ≤ 100, 2 ≤ t ≤ 10) — the length of the number and the number it should be divisible by.

Output

Print one such positive number without leading zeroes, — the answer to the problem, or  - 1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.

Sample Input

Input

3 2

Output

712

Source

Codeforces Round #324 (Div. 2)

//题意：

给你一个n，t，让你输出一个可以被t整除的一个n位数。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#define INF 0x3f3f3f3f
#define ull unsigned lonb long
#define ll long long
#define IN __int64
#define N 110
#define M 1000000007
using namespace std;
int main()
{
	int n,t;
	while(scanf("%d%d",&n,&t)!=EOF)
	{
		
		if(t<10&&t>=2)
		{
			for(int i=1;i<=n;i++)
				printf("%d",t);
		}
		else if(t==10)
		{
			if(n!=1)
			{
				for(int i=1;i<n;i++)
					printf("1");
				printf("0");
			}
			else
				printf("-1");	
		}
		else
			printf("-1");
		printf("\n");
	}
	return 0;
}