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;

typedef long long int ll;

int main(){
	int n,m;cin >> n >> m;
	vector<int> arr(n + 1, -1);
	while(m--){
		int s,c;cin >> s >> c;
		if(arr[s] == -1 || arr[s] == c)arr[s] = c;
		else return cout << -1, 0;
	}
	if(arr[1] == 0 && n > 1)return cout << -1, 0;
	for(int i = 1;i <= n;i++){
		if(i == 1 && arr[i] == -1 && n > 1)cout << 1;
		else if(arr[i] == -1)cout << 0;
		else cout << arr[i];
	}
}
