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,d[5],c,s;

int main()
{
	scanf("%d %d",&n,&m);
	for(int i=1;i<=n;++i) d[i]=-1;
	for(int i=1;i<=m;++i) {
		scanf("%d %d",&c,&s);
		if(d[c]!=-1&&d[c]!=s) {
			printf("-1");
			return 0;
		}
		d[c]=s;
	}
	if(d[1]==0&&n!=1) {
		printf("-1");
		return 0;
	}
	if(n==1&&d[1]==-1) {
		printf("0");
		return 0;
	}
	int ans=0;
	for(int i=1;i<=n;++i) {
		if(i==1&&d[i]==-1) {
			ans=1;continue;
		}
		if(d[i]!=-1) ans=ans*10+d[i];
		else ans=ans*10;
	}
	printf("%d",ans);
}