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) {

        System.out.println(solve());
    }

    private static int solve() {
        Scanner sc = new Scanner(System.in);

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

        int[] a = new int[N];
        Arrays.fill(a, -1);

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

            if (a[s] != -1 && a[s] != c) {
                return -1;
            }

            a[s] = c;
        }

        if (a[0] == 0) {
            if (N == 1) {
                return 0;
            } else {
                return -1;
            }

        } else {
            if (a[0] == -1) {
                if (N == 1) {
                    return 0;
                }
                a[0] = 1;
            }

            for (int i = 1; i < N; i++) {
                if (a[i] == -1) {
                    a[i] = 0;
                }
            }

            int ans = 0;
            for (int i = 0; i < N; i++) {
                ans = ans * 10;
                ans += a[i];
            }

            return ans;
        }

    }

}
