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 {
    Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {
        Main m = new Main();
        m.run();
    }

    private void run() {
        int n = sc.nextInt();
        int m = sc.nextInt();

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

        for (int i = 0; i < m; i++) {
            int s = sc.nextInt();
            int c = sc.nextInt();
            if (a[s] != -1 && a[s] != c || s == 1 && c == 0 && n > 1) {
                System.out.println(-1);
                return;
            }
            a[s] = c;
        }

        StringBuilder sb = new StringBuilder();

        if (n > 1) {
            sb.append(a[1] == -1 ? 1 : a[1]);
        } else {
            sb.append(a[1] == -1 ? 0 : a[1]);
        }
        for (int i = 2; i < n+1; i++) {
            sb.append(a[i] == -1 ? 0 : a[i]);
        }
        System.out.println(sb.toString());
    }
}
