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;
import java.lang.Math;

public class Main {
  public static void main(String[] args) {
    Scanner scan = new Scanner(System.in);
    int N = scan.nextInt();
    int M = scan.nextInt();
    int[] s = new int[M];
    int[] c = new int[M];
    for(int i=0; i<M; i++) {
      s[i] = scan.nextInt();
      c[i] = scan.nextInt();
    }
    scan.close();

    int min = N==1 ? 0 : (int)Math.pow(10, N-1);
    int max = (int)Math.pow(10, N) - 1;
    for(int i=min; i<=max; i++) {
      boolean test = true;
      for(int j=0; j<M; j++) {
        if((i % (int)Math.pow(10, N-s[j]+1)) / (int)Math.pow(10, N-s[j]) != c[j]) {
          test = false;
        }
      }
      if(test == true) {
        System.out.println(i);
        return;
      }
    }
    System.out.println(-1);
  }
}