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 N 5
int n,m,ans[N];
int main()
{
	scanf("%d %d",&n,&m);
	for(int i=1;i<=n;++i) ans[i]=-1;
	for(int i=1,x,y;i<=m;++i)
	{
		scanf("%d %d",&x,&y);
		if(~ans[x] && ans[x]!=y) {puts("-1");return 0;}
		ans[x]=y; 
	}
	if(!ans[1] && n>1) {puts("-1");return 0;}
	if(~ans[1]) printf("%d",ans[1]);
	else if(n==1) putchar('0');else putchar('1');
	for(int i=2;i<=n;++i)
		if(ans[i]==-1) putchar('0');
		else printf("%d",ans[i]);return 0; 
}