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, s, c;
	int num[3] = { 0 };
	vector<bool> check(3, true);
	bool out = false;

	cin >> n >> m;
	if (n == 3)num[2] = 1;
	if (n == 2)num[1] = 1;

	for (int i = 0; i < m; i++) {
		cin >> s >> c;
		if (n >=2 && s == 1 && c == 0)out = true;
		if (check.at(n - s) || num[n - s] == c) {
			num[n - s] = c;
			check.at(n - s) = false;
		}else out = true;
	}

	
	if (!out)cout << 100 * num[2] + 10 * num[1] + num[0] << endl;
	else cout << -1 << endl;
}
