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 void main(String[] args) {
		Main main = new Main();
		main.run();
	}

	public void run() {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		int a[] = new int[n+1];
		for(int i=0; i<=n; i++) {
			a[i]=-1;
		}
		for(int i=0; i<m; i++) {
			int s = sc.nextInt();
			int c = sc.nextInt();
			if(a[s]!=-1 && a[s]!=c) {
				System.out.println(-1);
				sc.close();
				return;
			}
			a[s]=c;
		}
		int ans = 0;
		for(int i=1; i<=n; i++) {
			int d=-1;
			if(i==1) {
				if(n>1) {
					if(a[i]==0) {
						System.out.println(-1);
						sc.close();
						return;
					}else if(a[i]==-1) {
						d=1;
					} else {
						d=a[i];
					}
				}else {
					if(a[i]==-1) {
						d=0;
					}else {
						d=a[i];
					}
				}
			}else {
				if(a[i]==-1) {
					d=0;
				}else {
					d=a[i];
				}
			}
			ans = ans * 10 + d;
		}
		System.out.println(ans);
		sc.close();
	}
}
