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 scan = new Scanner(System.in);
		
		int n = scan.nextInt();
		int m = scan.nextInt();
		int dn[] = new int[n];
		int ans = 0;
		
		for(int i=0;i<n;i++) {
			dn[i] = -1;
		}
		for(int i=0;i<m;i++) {
			int s = scan.nextInt();
			int c = scan.nextInt();
			if((dn[s-1] != -1 && dn[s-1] != c) || (s==1 && n!=1 && c==0)) {
				System.out.println(-1);
				System.exit(0);
			}else {
				dn[s-1] = c;				
			}
		}
		
		for(int i=0;i<n;i++) {
			if(dn[i] == -1 && (i==0 && n!=1)) {
				dn[i] = 1;
			}else if(dn[i] == -1) {
				dn[i] = 0;
			}
			
			ans += dn[i] * Math.pow(10,n-1-i);
		} 
		
		System.out.println(ans);
		
	}
	
}


