【Codeforces Round 324 (Div 2)A】【水题】Olesya and Rodion 构造数长度为n且是t的倍数

A. Olesya and Rodion

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 test(s)

input

3 2

output

712

#include<stdio.h>#include<string.h>#include<ctype.h>#include<math.h>#include<iostream>#include<string>#include<set>#include<map>#include<vector>#include<queue>#include<bitset>#include<algorithm>#include<time.h>using namespace std;void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}#define MS(x,y) memset(x,y,sizeof(x))#define MC(x,y) memcpy(x,y,sizeof(x))#define MP(x,y) make_pair(x,y)#define ls o<<1#define rs o<<1|1typedef long long LL;typedef unsigned long long UL;typedef unsigned int UI;template <class T1,class T2>inline void gmax(T1 &a,T2 b){if(b>a)a=b;}template <class T1,class T2>inline void gmin(T1 &a,T2 b){if(b<a)a=b;}const int N=0,M=0,Z=1e9+7,ms63=1061109567;int casenum,casei;int main(){int n,t;while(~scanf("%d%d",&n,&t)){if(t<10){printf("%d",t);for(int i=2;i<=n;i++)printf("0");puts("");}else{if(n==1)puts("-1");else{printf("%d",t);for(int i=3;i<=n;i++)printf("0");puts("");}}}return 0;}/*【题意】让你构造一个数。长度为n（1<=n<=100），且这个数是t（2<=t<=10）的倍数。【类型】水题【分析】因为t只有两种情况——1，t<10，即t为个位数。这个时候我们直接在后面添0即可。2，t=10，这个时候当n=1时非法，否则还是在后面添0即可。*/