If there is an integer not less than 0 satisfying the following conditions, print the smallest such integer; otherwise, print `-1`.

* The integer has exactly N digits in base ten. (We assume 0 to be a 1-digit integer. For other integers, leading zeros are not allowed.)
* The s_i-th digit from the left is c_i. \left(i = 1, 2, \cdots, M\right)

Constraints

* All values in input are integers.
* 1 \leq N \leq 3
* 0 \leq M \leq 5
* 1 \leq s_i \leq N
* 0 \leq c_i \leq 9

Input

Input is given from Standard Input in the following format:


N M
s_1 c_1
\vdots
s_M c_M


Output

Print the answer.

Examples

Input

3 3
1 7
3 2
1 7


Output

702


Input

3 2
2 1
2 3


Output

-1


Input

3 1
1 0


Output

-1
#include<iostream>
#include<bits/stdc++.h>
using namespace std;
int main()
{
	int N,M,s[4],c[10];
	cin>>N>>M;
	if(N==1&&M==0)
	{
		cout<<0;
		return 0;
	}
	int fla=1;
	memset(s,-1,sizeof(s));
	for(int i=1;i<=M;i++)
	{
		int ss,cc;
		cin>>ss>>cc;
		if(N>1&&ss==1&&cc==0)fla=0;
		else if(s[ss]==-1)s[ss]=cc;
		else  
		{
			if(s[ss]!=-1&&s[ss]!=cc)fla=0;
		}
	}
	if(fla==0)
	{
		cout<<"-1";
		return 0;
	}
	for(int i=1;i<=N;i++)
	{
		if(s[i]==-1&&i==1)s[i]=1;
		else if(s[i]==-1)s[i]=0;
		printf("%d",s[i]);
	}
	return 0;
	
} 