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;
typedef long long  ll;
int main(){
  int n,m;
  cin>>n>>m;
  vector<int> s(n);
  bool ok = true;
  for(int i=0; i<m; i++){
    int tmp1, tmp2;
    cin>>tmp1>>tmp2;
    if(s.at(tmp1-1)!=0&&s.at(tmp1-1)!=tmp2) ok =false;
    else s.at(tmp1-1) =tmp2;
    if(tmp1==1&&tmp2==0&&n>1) ok = false;
  }
  if(ok&&s.at(0)==0&&n>1) s.at(0)=1;
  if( ok ){
    for(int i =0; i<n; i++){
      cout<<s.at(i);
    }
    cout<<endl;
  }
  else cout<<-1<<endl;
}
