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,y,z;
  cin >> n >> m;
  vector<int> x(n,10);
  for (int i = 0; i < m; i++) {
    cin >> y >> z;
    if(y==1&&z==0&&n!=1) {
      cout << -1 << endl;
      exit(0);
    }
    if(x[y-1]!=10&&x[y-1]!=z) {
      cout << -1 << endl;
      exit(0);
    }
    x[y-1]=z;
  }
  if(x[0]==10) {
    if(n!=1){
      x[0]=1;
    }else{
      x[0]=0;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << x[i]%10;
  }
  cout << endl;
}