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.*;
public class Main{
  public static int n,m;
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    n = sc.nextInt();
    m = sc.nextInt();
    int[] s = new int[m];
    int[] c = new int[m];
    for(int i=0;i<m;i++){
      s[i] = sc.nextInt();
      c[i] = sc.nextInt();
    }
    
    System.out.println(solve(s,c));
  }
  
  public static String solve(int[] s,int[] c){
    for(int i=0;i<1000;i++){
      String str=String.valueOf(i);
      if(ok(str,s,c)) return str; 
    }
    return "-1";
  }
  
  public static boolean ok(String str,int[] s,int[] c){
    if(n!=str.length()) return false;
    for(int i=0;i<m;i++){
      if(str.charAt(s[i]-1)!=c[i]+'0') return false;
    }
    return true;
  }
  
}