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;
	cin >> n >> m;
	vector<int> x(n, -1);	
	for (int i = 0; i < m; i++) {
		int s, c;
		cin >> s >> c;
		s--;
		if (x[s] != -1 && x[s] != c) {
			cout << -1 << endl;
			return 0;
		}
		x[s] = c;
	}
	if (n > 1 && x[0] == 0) {
		cout << -1 << endl;
		return 0;
	}
	if (n > 1 && x[0] == -1) x[0] = 1;
	int ans = 0;
	for (int i = 0; i < n; i++) {
		ans *= 10;
		ans += max(0, x[i]);
	}
	cout << ans << endl;
	return 0;
}