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> A(N);
  for(int i=0;i<N;i++){
    A[i] = 0;
  }
  for(int i=0;i<M;i++){
    int a, b;
    cin >> a >> b;
    if(a == 1&&b==0 && N!= 1){
      cout << -1 << endl;
      return 0;
    }
    if(A[a-1] != 0 && A[a-1] != b){
      cout << -1 << endl;
      return 0;
    }else{
      A[a-1] = b;
    }
  }
  if(A[0] == 0 && N!=1)A[0] = 1;
  for(int i=0;i<N;i++){
    cout << A[i];
  }
  cout << endl;
}
  
