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;

public class Main {
	static String s = "";

	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ
		Scanner sc = new Scanner(System.in);
		int ans = 0;

		int N = sc.nextInt();
		int M = sc.nextInt();

		int[] kurai = new int[N];
		for(int i=0; i<N; i++) {
			kurai[i] = -1;
		}

		for(int i=0; i<M; i++) {
			int s = sc.nextInt();
			int c = sc.nextInt();

			if(kurai[s-1] == -1) {
				kurai[s-1] = c;
			}else {
				if(kurai[s-1] != c) {
					ans = -1;
					break;
				}
			}
		}


		if(N == 1) {
			if(kurai[0] == -1) {
				kurai[0] = 0;
			}
		}
		else {
			for(int i=0; i<N; i++) {
				if(kurai[i] == -1) {
					if(i == 0)
						kurai[i] = 1;
					else
						kurai[i] = 0;
				}
			}
		}

		if(ans == -1 || (N != 1 && kurai[0] == 0))
			ans = -1;
		else {
			int j = 0;
			for(int i=N-1; i>=0; i--) {
				ans += kurai[i] * Math.pow(10, j);
				j++;
			}
		}

		System.out.println(ans);

	}

}
