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 sc = new Scanner(System.in);
		int N = sc.nextInt();
		int M = sc.nextInt();
		int flg2 = 0;
		int[] c = new int[N];
		int[] flg = new int[N];
		int[] bufn = new int[N];
		int i,s,j,dust;
		int buf;
		int ans = -1;
		for(i = 0;i < M;i++){
			s = sc.nextInt();
			c[s-1] = sc.nextInt();
			if(flg[s-1] == 0){
				bufn[s-1] = c[s-1];
				flg[s-1] = 1;
			}else if(bufn[s-1] != c[s-1]){
				flg2 = 1;
			}
			if(N != 1 && flg[0] == 1 && c[0] == 0){
				flg2 = 1;
			}
		}
		if(flg2 == 0){
			ans = 0;
			for(i = 0;i < N;i++){
				buf = 1;
				for(j = 0;j < i;j++){
					buf *= 10;
				}
				ans += (c[N-1-i] * buf);
				if(N != 1 && i == (N-1) && c[0] == 0){
					ans += buf;
				}
			}
		}
		System.out.println(ans);
	}
}