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,p,tmp,j=0;
  cin >> N >> M;
  vector<int> num(N,-1);
  for(int i=0;i<M;i++) {
    cin >> p >> tmp;
    if(num.at(p-1)==-1||num.at(p-1)==tmp) {
      num.at(p-1)=tmp;
    } else{
      j--;
    }
  }  
  if(N!=1&&num.at(0)==0||j<0){
    cout << -1 << endl;
  } else {
    for(int i=0;i<N;i++) {
      if(N!=1&&i==0&&num.at(i)==-1) cout << 1;
      else if(num.at(i)==-1) cout << 0;
      else cout << num.at(i);
    }
    cout << endl;
  }
}
