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.*;
import java.io.PrintWriter;
public class Main{
	public static void main(String[] args){
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		int m=sc.nextInt();
		if(n==1 && m==0){
			System.out.println(0);
			return;
		}
		StringBuilder ans=new StringBuilder();
		int[] s=new int[n];
		Arrays.fill(s,-1);
		for(int i=0; i<m; i++){
			int tmps=sc.nextInt()-1;
			int tmpc=sc.nextInt();
			if(n!=1 && tmps==0 && tmpc==0){
				System.out.println(-1);
				return;
			}
			if(s[tmps]!=-1 && s[tmps]!=tmpc){
				System.out.println(-1);
				return;
			}
			s[tmps]=tmpc;
		}
		for(int i=0; i<n; i++){
			if(s[i]!=-1){
				ans.append(s[i]);
			}else{
				if(i==0){
					ans.append("1");
				}else{
					ans.append("0");
				}
			}
		}
		System.out.println(ans);
	}
}
