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.*;
import java.math.*;

public class Main {

  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int 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();
    }

    TreeMap<Integer,Integer> mp = new TreeMap<Integer,Integer>();

    for (int i = 0; i < m; i++) {
      if (mp.containsKey(s[i])) {
        int k = mp.get(s[i]);
        if (k != c[i]) {
          System.out.println(-1);
          return;
        }
      }
      mp.put(s[i],c[i]);
    }

    String h = "";

    for (int i = 1; i <= n; i++) {
      Integer x = mp.get(i);
      if (x == null) {
        if (i == 1 && n != 1) {
          h += "1";
        } else {
          h += "0";
        }
      } else {
        h += String.valueOf(x);
      }
    }

    if (h.length() > 1 && h.charAt(0) == '0') {
      System.out.println(-1);
    } else {
      System.out.println(h);
    }

  }
}
