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 ;
#define pb push_back
#define endl '\n'
#define x first
#define y second

int main(){
	int n,m,s,c;
	int a[4]={-1,-1,-1,-1};
	cin>>n>>m;
	bool f=0;
	while(m--){
		cin>>s>>c;
		s--;
		if(a[s]!=-1){
			if(a[s]!=c)f=1;
		}
		a[s]=c;
	}
	if(f){
		cout<<-1<<endl;
		return 0;
	}
	if(n!=1 &&a[0]==0){
		cout<<-1<<endl;
		return 0;
	}
	if(a[0]==-1){
		if(n==1){
			a[0]=0;
		}else{
			a[0]=1;
		}
	}
	for(int i=0;i<n;i++){
		if(a[i]==-1){
			a[i]=0;
		}
		cout<<a[i];
	}
	cout<<endl;
}