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
import java.util.*;

public class Main{
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
  	int n,m;
  	
  	n=sc.nextInt();
  	m=sc.nextInt();
  	
  	int[] s=new int[m];
  	char[] c=new char[m];
  	for(int i=0;i<m;i++){
  		s[i]=sc.nextInt();
  		c[i]=sc.next().charAt(0);
  	}
  	int f=0,g=0;
  	if(n==1){
  		for(int i=0;i<10;i++){
  			int e=0;
  			for(int j=0;j<m;j++){
  				String is=Integer.toString(i);
  				char[] ic=is.toCharArray();
  				if(ic[s[j]-1]==c[j]){
  					e++;
  				}else{
  					e=0;
  				}
  			}
  			if(e==m){
  				f++;
  				g=i;
  				break;
  			}
  		}
  	}else{
  		for(int i=(int)(Math.pow(10,n-1));i<(int)(Math.pow(10,n));i++){
  			int e=0;
  			for(int j=0;j<m;j++){
  				String is=Integer.toString(i);
  				char[] ic=is.toCharArray();
  				if(ic[s[j]-1]==c[j]){
  					e++;
  				}else{
  					e=0;
  				}
  			}
  			if(e==m){
  				f++;
  				g=i;
  				break;
  			}
  		}
  	}
  	if(f==0){
  		System.out.print(-1);
  	}else{
  		System.out.print(g);
  	}
  }
}