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.Arrays;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int m = scanner.nextInt();
        Integer[] digits = new Integer[n];
        Arrays.fill(digits, null);
        for (int i = 0; i < m; i++) {
            int s = scanner.nextInt()-1;
            int c = scanner.nextInt();
            if(digits[s]==null){
                digits[s]=c;
            }else{
                if(digits[s]!=c){
                    System.out.println(-1);
                    return;
                }
            }
        }
        if(digits[0]==null){
            digits[0] = n==1? 0 : 1;
        }
        if(digits[0]==0 && n!=1){
            System.out.println(-1);
            return;
        }
        for (int i = 1; i < n; i++) {
            if(digits[i]==null){
                digits[i]=0;
            }
        }
        int res = digits[0];
        for (int i = 1; i < n; i++) {
            res = res*10+digits[i];
        }
        System.out.println(res);
    }
}