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 {
 
	public static final MyScanner in = new MyScanner();
 
	public static void main(String[] args) {
 
		int N  = in.nextInt();
		int M = in.nextInt();
		int[] answer = new int[N];
		for(int i=0;i<N;i++) {
			answer[i] = -1;
		}

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

			// 一桁目が0
			if (N != 1 && s == 1 && c == 0) {
				System.out.println(-1);
				return;
			}

			// 矛盾した定義
			int a = answer[s-1];
			if(a != -1) {
				if (a != c) {
					System.out.println(-1);
					return;
				}
			}

			answer[s-1] = c;
		}

		if (answer[0] == -1 && N != 1) answer[0] = 1;
		if (answer[0] == -1 && N == 1) answer[0] = 0;
		for(int i=1;i<N;i++) {
			if (answer[i] == -1) answer[i] = 0;
		}

		for(int i=0;i<N;i++) {
			System.out.print(answer[i]);
		}
		System.out.println();
	}
 
 
	public static class MyScanner{
		Scanner sc = new Scanner(System.in);
		String nextLine() {return sc.nextLine();}
		String next() {return sc.next();}
		int nextInt() {return Integer.valueOf(sc.next());}
		long nextLong() { return Long.valueOf(sc.next());}
	}

}