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<pair<int, int>> p(m);
	for (int i = 0; i < m; i++) {
		cin >> p[i].first >> p[i].second;
		p[i].first--;
	}
	for (int i = 0; i <= 999; i++) {
		string s = to_string(i);
		if ((int)s.size() < n) continue;
		bool ok = true;
		for (auto a : p) {
			if (s[a.first] != a.second + '0') ok = false;
		}
		if (ok) {
			cout << s << endl;
			return 0;
		}
	}
	cout << -1 << endl;
	return 0;
}