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) throws Exception {
        try (Scanner sc = new Scanner(System.in)) {
            int N = sc.nextInt();
            int M = sc.nextInt();

            char[] C = new char[N];
            Arrays.fill(C, '#');

            for (int i = 0; i < M; i++) {
                int s = sc.nextInt() - 1;
                char c = (char) (sc.nextInt() + '0');

                if (C[s] != '#' && C[s] != c) {
                    System.out.println(-1);
                    return;
                }

                C[s] = c;
            }

            if (C[0] == '0' && N > 1) {
                System.out.println(-1);
                return;
            }

            if (C[0] == '#' && N == 1) {
                System.out.println(0);
                return;
            }

            for (int i = 0; i < N; i++) {
                if (C[i] == '#') {
                    if (i == 0) {
                        C[i] = '1';
                    } else {
                        C[i] = '0';
                    }
                }
            }

            System.out.println(String.valueOf(C));
        }
    }

}
