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
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		int n = in.nextInt();
		int m = in.nextInt();
		int[] s = new int[m];
		int[] c = new int[m];
		for (int i = 0; i < m; i++) {
			s[i] = in.nextInt() - 1;
			c[i] = in.nextInt();
		}
		int r = solve(n, m, s, c);
		System.out.println(r);
		in.close();
	}

	private static int solve(int n, int m, int[] s, int[] c) {
		int[] r = new int[n];
		for (int i = 0; i < n; i++)
			r[i] = -1;
		for (int i = 0; i < m; i++)
			if (r[s[i]] == -1)
				r[s[i]] = c[i];
			else if (r[s[i]] != c[i])
				return -1;
		if (n > 1 && r[0] == 0)
			return -1;
		if (r[0] == -1)
			if (n > 1)
				r[0] = 1;
			else
				r[0] = 0;
		for (int i = 1; i < n; i++)
			if (r[i] == -1)
				r[i] = 0;
		int ans = 0;
		for (int i = 0; i < n; i++) {
			ans *= 10;
			ans += r[i];
		}
		return ans;
	}

}
