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;
int s[6],c[6];
bool v[6];
int t[10];
int main() {
	memset(v,0,sizeof(v));
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++) {
		scanf("%d%d",&s[i],&c[i]);
		if(v[s[i]] && t[s[i]]!=c[i]) {
			cout<<-1;
			exit(0);
		}
		v[s[i]]=1;
		t[s[i]]=c[i];
	}
	if(n==1 && t[1]==0) {
		printf("0\n");
		exit(0);
	}
	if(v[1]==0) {
		t[1]=1;
	}
	bool first=1;
	for(int i=1;i<=n;i++) {
		if(first && t[first]==0) {
			cout<<-1;
			exit(0);
		}
		first=0;
		printf("%d",t[i]);
	}
	return 0;
}