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=sc.nextInt();
		int m=sc.nextInt();
		int ans =-1;
		int[] digits =new int[n];
		int[] s_array=new int[m];
		int[] c_array=new int[m];
		
		for (int i=0;i<m;i++) {
			s_array[i]=sc.nextInt();
			c_array[i]=sc.nextInt();
		}
		int start =0;
		if(n==2) {
			start=10;
		}
		if(n==3) {
			start =100;
		}
		for (int i=start;i<1000;i++) {
			int i_digit =i;
			for (int j=n-1;j>=0;j--) {
				digits[j]=i_digit%10;
				i_digit /= 10;
			}

			boolean isAns=true;

			for (int k=0;k<m;k++) {
				if (digits[s_array[k]-1]!=c_array[k]) {
					isAns =false;
				}
			}
			if (isAns) {
				ans=i;
				break;
			}
		}
		//━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
		System.out.println(ans);
	}
	//----------------------------------------
	//メソッド置き場

	//----------------------------------------
}