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 N,M;
vector<int> s,c;

bool ok(string str){
  if(str.size() !=N) return false;
  for(int i=0;i<M;i++) if(str[s[i]]!=(char)(c[i]+'0')) return false;
  return true;
}

string solve() {
  for(int i=0;i<1000;i++){
    stringstream iss; iss << i;
    string str = iss.str();
    if(ok(str)) return str;
  }
  return "-1";
}
int main() {
  cin>> N >> M;
  s.resize(M); c.resize(M);
  for(int i=0;i<M;i++) cin >> s[i] >> c[i], s[i]--;
  cout << solve() << endl;
}