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 ;
	vector<pair<int,int>>vi;
	cin>>n>>m;
	for(int i = 0; i < m ; i++){
		int a , b;
		cin>>a>>b;
		a--;
		vi.push_back({a,b});
	}
	for(int i = 0; i < 1000 ; i++){
		string s = to_string(i);
		if((int)s.size() != n)continue;
		bool can = true;
		for(int j = 0; j < m ; j++){
			if(s[vi[j].first] != (char)vi[j].second + '0'){
				can = false;
			}
		}
		if(can)return cout<<s,0;
	}
	cout<<-1;
	return 0;
}

