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 main() {
	int n, m; scanf("%d%d",&n,&m);
	int d[n+1];
	for (int i=1; i<=n; i++) d[i]=10;
	while (m--) {
		int a, b; scanf("%d%d",&a,&b);
		if (d[a]!=10 && d[a]!=b) {
			printf("-1\n");
			return 0;
		}
		if (a==1 && b==0 && n>1) {
			printf("-1\n");
			return 0;
		}
		d[a]=b;
	}
	if (d[1]==10 && n>1) d[1]=1;
	for (int i=1; i<=n; i++) {
		printf("%d",d[i]==10?0:d[i]);
	}
	return 0;
}