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.Scanner;

class Main {
	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);

		int N = Integer.parseInt(sc.next());
		int M = Integer.parseInt(sc.next());
//		String S = sc.next();

		int[] arr = new int[N];
		for(int i = 0; i< N ; i++) {
			arr[i] = 10;
		}
		for(int i = 0; i<M ; i++) {
			int aa = Integer.parseInt(sc.next());
			int bb = Integer.parseInt(sc.next());

			if(N > 1 && aa ==1 && bb == 0) {
				System.out.println(-1);
				return;
			}
			if(arr[aa-1] != 10 && arr[aa-1] != bb) {
				System.out.println(-1);
				return;
			} else {
				arr[aa-1] =  bb;
			}
		}


		for(int i=0;i < arr.length;i++) {
			if(arr.length > 1 && i==0 && arr[i] == 10) {
				System.out.print(1);
			} else if(arr[i] == 10) {
				System.out.print(0);
			} else {
				System.out.print(arr[i]);
			}
		}
	}
}
