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;

const int M=10;

int nu[M];

int main()
{
	int n,m; scanf("%d%d",&n,&m);
	int s,c;
	memset(nu,-1,sizeof(nu));
	for(int i=0;i<m;i++)
	{
		scanf("%d%d",&s,&c);
		if(nu[s]==-1||nu[s]==c)
			nu[s]=c;
		else
		{
			printf("-1\n");
			return 0;
		}
	}
	if(nu[1]==-1)
		nu[1]=n==1?0:1;
	if(nu[2]==-1) nu[2]=0;
	if(nu[3]==-1) nu[3]=0;
	if(n!=1&&nu[1]==0)
	{
		printf("-1\n");
		return 0;
	}
	for(int i=1;i<=n;i++)
		printf("%d",nu[i]);
	putchar('\n');
	return 0;
}