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 sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        int[] s = new int[m];
        int[] c = new int[m];

        for (int i = 0; i < m; i++) {
            s[i] = sc.nextInt();
            c[i] = sc.nextInt();
        }

        int[] num = new int[n + 1];
        Arrays.fill(num, -1);
        for (int i = 0; i < m; i++) {
            if (num[s[i]] != -1 && num[s[i]] != c[i]) {
                System.out.println(-1);
                return;
            }
            num[s[i]] = c[i];
        }
        for (int i = 1; i < num.length; i++) {
            if (i == 1 && num[i] == 0 && i + 1 < num.length) {
                System.out.println(-1);
                return;
            }
            if (num[i] == -1) {
                if (i == 1 && i + 1 < num.length) {
                    num[i] = 1;
                    continue;
                }
                num[i] = 0;
            }
        }
        int ans = 0;
        for (int i = 1; i < num.length; i++) {
            ans += num[i] * Math.pow(10, num.length - i - 1);
        }
        System.out.println(ans);
    }

}
