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 ll long long
#define bug(x) cerr<<#x<<"="<<x<<endl;



int main( ){
  int n,m,pos,num;

  scanf("%d %d",&n,&m);

   vector<int>s(n+1,-1);

  bool ex=0;

  for(int i=0;i<m;i++){

    scanf("%d %d",&pos,&num);
    pos--;

    if((s[pos]>-1&&s[pos]!=num)||(pos+num<1&&n>1)){
        return cout<<-1<<endl,0;
    }
    s[pos]=num;
  }

    cout<<(s[0]==-1?(n>1?1:0):s[0]);
    for(int i=1;i<n;i++)cout<<max(s[i],0);
    cout<<endl;

}