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.Scanner;
public class Main {

	public static void main(String[] args) {
		Scanner stdIn = new Scanner(System.in);
		
		int n = stdIn.nextInt(); //n桁
		int m = stdIn.nextInt();
		
		boolean[] flag = new boolean[n];
		
		int[] check = new int[n]; //ある桁の場所に同じ数値が入ったかどうかcheck
		
		
		for(int i = 0; i < m; i++) {
			int s = stdIn.nextInt()-1;
			int c = stdIn.nextInt();
			
			if(s==0 && c==0 && n!=1) {
				System.out.println(-1);
				return;
			}
			
			if(flag[s]==true && check[s]!=c) {
				System.out.println(-1);
				return;
			}
			
			flag[s] = true;
			check[s] = c;
			
		}
		
		if(n==1) {
			if(flag[0]==false ) {
				System.out.println(0);
				return;
			}else {
				System.out.println(check[0]);
				return;
			}
		}
		
		for(int i = 0; i < n; i++) {
			if(flag[i]==false && i==0) {
				System.out.print(1);
			}else if(flag[i] ==false) {
				System.out.print(0);
			}else {
				System.out.print(check[i]);
			}
		}
		
	}

}
