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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
int s[6],t[6];
int a[3]={-1,-1,-1};
int main(void){
	int n,m; cin>>n>>m;
	for(int i=0;i<m;i++){
		cin>>s[i]>>t[i];
		if(a[s[i]-1]!=-1&&a[s[i]-1]!=t[i]){
			cout<<-1<<endl;
			return 0;
		}else{
			a[s[i]-1]=t[i];
		}
	}
	if(a[0]==0&&n!=1){
		cout<<-1<<endl;
		return 0;
	}
	if(n==1&&m==0){
		cout<<0<<endl;
		return 0;
	}
	for(int i=0;i<n;i++){
		if(a[i]!=-1){
			cout<<a[i];
		}else if(i==0){
			cout<<1;
		}else{
			cout<<0;
		}
	}
	cout<<endl;
	    
}
