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) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int[] a = new int[n];
    int[] d = new int[n];
    Arrays.fill(d, -1);
    int m = sc.nextInt();
    for (int i = 0; i < m; i++) {
      int si = sc.nextInt() - 1;
      int ci = sc.nextInt();
      if(d[si] != -1 && d[si] != ci){
        System.out.println(-1);
        return;
      }
      if(n != 1 && si == 0 && ci == 0){
        System.out.println(-1);
        return;
      }
      d[si] = ci;
      a[si] = ci;
    }
    if(d[0] == -1 && n != 1){
      a[0] = 1;
    }
    int answer = 0;
    int pow = 1;
    for(int i = 0; i < n; i++){
      answer += a[n - 1 - i] * pow;
      pow *= 10;
    }
    System.out.println(answer);
  }
}