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.Scanner;

public class Main {
    static int N;
    static int M;
    static int[] s;
    static int[] c;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        N = sc.nextInt();
        M = sc.nextInt();
        s = new int[M];
        c = new int[M];
        for(int i = 0; i < M; i++) {
            s[i] = sc.nextInt();
            c[i] = sc.nextInt();
            --s[i];
        }
        System.out.println(solve());
    }

    static String solve() {
        for(int i = 0; i <= 999; i++) {
            String str = String.valueOf(i);
            if(ok(str)) return str;
        }
        return "-1";
    }

    static boolean ok(String str) {
        if(N != str.length()) return false;
        for(int i = 0; i < M; i++) {
            if(str.charAt(s[i]) != (char)(c[i]+'0')) return false;
        }
        return true;
    }
}
