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 N,M;
  cin >> N >> M;
  vector<int> V(N,-1);
  
  for(int i=0;i<M;i++){
    int s,c;
    cin >> s >> c;
    s--;
    if(V[s]==-1) V[s] = c;
    else if(V[s]!=c){
      cout << -1 << endl;
      return 0;
    }
  }
  
  if(N==1) cout << max(V[0],0) << endl;
  else{
    if(V[0]==0){
      cout << -1 << endl;
      return 0;
    }
    else cout << max(V[0],1);
    for(int i=1;i<N;i++) cout << max(V[i],0);
    cout << endl;
  }
  return 0;
}
