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 sc = new Scanner(System.in);

		int n = Integer.parseInt(sc.next());
		
		int []a = new int [n];
		boolean []b = new boolean [n];
		int m = Integer.parseInt(sc.next());
		
		boolean flag = true;
		
		for(int i = 0;i < m;i++) {
			int s = Integer.parseInt(sc.next());
			int t = Integer.parseInt(sc.next());
			if(!b[n-s]) {
				if(s == 1 && t == 0 && n >= 2) {
					flag = false;
					break;
				}
				a[n-s] = t; 
				b[n-s] = true;
			}
			else {
				if(a[n-s] != t) {
					flag = false;
					break;
				}
			}
		}
		
		if(flag) {
			int ans= 0;
			
			for(int i = 0 ; i < n ;i++) {
				if(i == n-1 && !b[i] && n >= 2) {
				ans += Math.pow(10, i);
				} 
				else {
				ans += a[i] * Math.pow(10, i);
			}
		}
			System.out.println(ans);
		}
		else {
			System.out.println(-1);
		}
	}

}
