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 <cstdio>
#include <cmath>
using namespace std;
int N, M, s[10], c[10];
int main() {
  scanf("%d %d", &N, &M);
  for(int i = 1; i <= M; ++i) scanf("%d %d", &s[i], &c[i]);
  int st = N>1?pow(10, N-1):0, ans = -1;
  for(int i = st; i <= pow(10, N)-1; ++i) {
    int num[3] = {i/100, (i%100)/10, (i%10)/1}, cnt = 0;
    for(int j = 1; j <= M; ++j) {
      if(num[s[j]-N+2] == c[j]) ++cnt;
    }
    if(cnt != M) continue;
    ans = i;
    break;
  }
  printf("%d\n", ans);
}