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.*;
class Main {
	public static void main(String[] args) throws Exception {
		final Scanner sc = new Scanner(System.in);

        final int N = sc.nextInt();
        final int M = sc.nextInt();

        int[] num = new int[N];
        for (int i = 0; i < num.length; i++) {
            num[i] = -1;
        }
        if (M == 0 && N == 1) {
            System.out.println(0);
            return;
        } else if (M != 0) {
            for (int i = 0; i < M; i++) {
                int keta = sc.nextInt() - 1;
                int value = sc.nextInt();
                if (num[keta] != -1 && num[keta] != value) {
                    System.out.println(-1);
                    return;
                } else if (keta == 0 && value == 0 && N != 1) {
                    System.out.println(-1);
                    return;
                }
                num[keta] = value;
            }
            sc.close();
        }

        for (int i = 0; i < num.length; i++) {
            if (num[i] == -1) {
                if (i == 0)
                    num[i] = 1;
                else
                    num[i] = 0;
            }
            System.out.print(num[i]);
        }

        System.out.println();
    }
}
