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 ans = 0;
        boolean error = false;
        int [] num = new int[N];
        for(int i=0;i<N;i++){
            num[i] = -1;
        }

        for(int i=0;i<M;i++){
            int s = sc.nextInt();
            int c = sc.nextInt();
            if(s > N){
                error = true;
                break;
            }
            if(c == num[s-1] || num[s-1] == -1){
                num[s-1] = c;
            } else {
                error = true;
                break;
            }

        }
        if(num[0] == 0){
            if(N != 1){
                error = true;
            }

        } else if (num[0] == -1){
            if(N != 1){
                num[0] = 1;
            } else {
                num[0] = 0;
            }
        }

        for(int i=0;i<N;i++) {
            if (num[i] == -1) {
                num[i] = i == 0 ? 1 : 0;
            }
            ans += num[i] * Math.pow(10, N - 1 - i);
        }

        if(error){
            ans = -1;
        }

        System.out.println(ans);


    }

}






