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{
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int[] array = new int[3];
    for(int i = 0; i < 3; i++){
      array[i] = -1;
    }
    int m = sc.nextInt();
    for(int i = 0; i < m; i++){
      int s = sc.nextInt();
      int c = sc.nextInt();
      if(n!=1&&s==1&&c==0){
        System.out.println(-1);
        return;
      }
      if(array[s-1]!=-1&&array[s-1]!= c){
        System.out.println(-1);
        return;
      }
      array[s-1] = c;
    }
    for(int i = 0; i < 3; i++){
      if(n!=1&&i==0&&array[i]==-1){
        array[i] = 1;
      }
      if(array[i]==-1){
        array[i]=0;
      }
    }
    int num = 0;
    for(int i = 0; i < n; i++){
      num = num*10;
      num+=array[i];
    }
    System.out.println(num);
  }
}