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.util.ArrayList;
import java.util.HashSet;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);

		int n = scan.nextInt();
		int m = scan.nextInt();
		boolean check = false;
		int num = 0;

		ArrayList<HashSet<Integer>> list = new ArrayList<HashSet<Integer>>(n);

		for(int i = 0;i < n;i++) {
			list.add(new HashSet<Integer>());
		}

		for(int i = 0;i < m;i++) {
			list.get(scan.nextInt() - 1).add(scan.nextInt());
		}

		if(m != 0) {
			for(int i = 0;i < n;i++) {
				if(list.get(0).contains(0)) {
					if(n != 1) {
						check = true;
						break;
					}
				}
				if(list.get(i).size() > 1) {
					check = true;
					break;
				}
			}

			if(check) {
				System.out.println(-1);
			}else {
				for(int i = 0;i < n ;i++) {
					if(list.get(i).isEmpty()) {
						if(i == 0) {
							num += 1 * Math.pow(10, n - 1);
						}

					}else {
						for (int a : list.get(i)) {
							num += a * Math.pow(10, n - 1 - i);
						}
					}
				}
				System.out.println(num);
			}
		}else {
			if(n == 1) {
				System.out.println(0);
			}else {
				num = (int) (1 * Math.pow(10, n - 1));
				System.out.println(num);
			}
		}
	}
}