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 a = sc.nextInt();
        int b = sc.nextInt();
        int c = 0;
		int [] in = new int [b];
        int [] str = new int [b];
        int [] ans = new int [a];
        for(int i=0;i<b;i++){str[i] = sc.nextInt();
                             in[i] = sc.nextInt();
                            }
        for(int i=0;i<a;i++){ans[i] = -1;
                            }
        for(int i=0;i<b;i++){if(ans[str[i]-1]==in[i]||ans[str[i]-1]==-1)
                              {ans[str[i]-1] = in[i];}
                              else{c = -1;}
                             if(a>1 && ans[0]==0){c=-1;}
                             }
		if(c==0){if(a==1) {c=Math.max(0,ans[0]);}
        else{for(int i=0;i<a;i++){if(ans[i]==-1){if(i==0){ans[i]=1;}else{ans[i]=0;}}
                                                                                                        
                                 }
               if(a==2){c=10*ans[0]+ans[1];} else{c=100*ans[0]+10*ans[1]+ans[2];}
            }
                }
        
		System.out.println(c);
	}
}
