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;
  cin >> n >> m;
  vector<int> a(n, -1);
  int ans = 0;
  for (int i = 0; i < m; i++) {
    int s, c;
    cin >> s >> c;
    s--;
    if (a[s] < 0 || a[s] == c) {
      a[s] = c;
    } else goto noans;
  }
  if (n > 1 && a[0] == 0) goto noans;
  ans = a[0] >= 0 ? a[0] : n > 1 ? 1 : 0;
  for (int i = 1; i < n; i++) {
    ans = ans * 10 + max(a[i], 0);
  }
  cout << ans << endl;
  return 0;
 noans:
  cout << -1 << endl;
}
