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;
void solve(){
  int n,m;
  cin >> n >> m;
  vector<int> s(m),c(m);
  for(int i=0;i<m;i++)cin >> s[i] >> c[i];
  for(int i=0;i<1000;i++){
    string x = to_string(i);
    if(x.size()==n){
      bool valid = true;
      for(int j=0;j<m;j++)if(x[s[j]-1]!=('0'+c[j]))valid=false;
      if(valid){cout << i << endl; return;}
    }
  }
  cout << -1 << endl;
}
int main(){
  solve();
  return 0;
}