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
#include <bits/stdc++.h>
using namespace std;

int main(){
  int a[5];
  a[0] = -1;
  a[1] = -1;
  a[2] = -1;
  int n,m;
  cin >> n >> m;
  int s,c;
  for(int i=0; i<m; i++){
    cin >> s >> c;
    if(a[n-s] == -1){
      a[n-s] = c;
    } else if(a[n-s] != c){
      cout << -1;
      return 0;
    }
  }
  if(n>=2 && a[n-1] == 0){
    cout << -1;
    return 0;
  } else if(n>=2 && a[n-1] == -1){
    a[n-1] = 1;
  }
  cout << max(a[0],0)+max(a[1],0)*10+max(a[2],0)*100;
}
