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;
#define rep(i, n) for(int i = 0; i < n; i++)

int main(){
  cin.tie(nullptr);ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(15);
  int n,m;cin>>n>>m;
  int s,c;
  vector<int> a(n,-1);
  rep(i,m){
    cin>>s>>c;
    if(!(n>=2 && s==1 && c==0)&&(a[s-1]==-1 ||a[s-1]==c)){
      a[s-1]=c;
    }else{
      cout<<-1<<endl;
      return 0;
    }
  }
  rep(i,n) n>=2 && i==0 && a[i]==-1 ? cout<<1 : a[i]!=-1 ? cout<<a[i] : cout<<0;
  cout<<endl;
}