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 rep(i,n) for(int i=0; i<(n); i++)
int n,m,s,c,a[3] = {-1,-1,-1};
bool ok = true;
int main(){
	scanf("%d%d", &n, &m);
	rep(i,m){
		scanf("%d%d", &s, &c);
		if(s == 1 && c == 0 && n > 1) ok = false;
		if(a[2-n+s] != -1){
			if(a[2-n+s] != c) ok = false;
		}else a[2-n+s] = c;
	}
	if(a[3-n] == -1) a[3-n] = n == 1 ? 0 : 1;
	if(ok){
		rep(i,n) printf("%d",a[3-n+i] == -1 ? 0 : a[3-n+i]);
		puts("");
	}else puts("-1");
}