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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;

public class Main {

	final static long MOD = 1000000007;
	public static void main(String[] args) throws Exception {
		// TODO 自動生成されたメソッド・スタブ

		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		String W = in.readLine();

		String[] Ws = W.split(" ");
		int N = Integer.parseInt(Ws[0]);
		int M = Integer.parseInt(Ws[1]);

		int[][] A = new int[M][2];
		for (int i = 0; i < M; i++) {
			W = in.readLine();
			Ws = W.split(" ");
			A[i][0] = Integer.parseInt(Ws[0]);
			A[i][1] = Integer.parseInt(Ws[1]);
		}

		int[] X = new int[N];
		Arrays.fill(X, -1);
		for(int i=0;i<A.length;i++){
			if((X[A[i][0]-1] == -1 || X[A[i][0]-1] == A[i][1]) ){
				X[A[i][0]-1] = A[i][1];
			}else{
				System.out.println(-1);
				return;
			}
		}

		int ans = 0;
		if(M!= 0 && X[0]==0 && N!=1){
			System.out.println(-1);
			return ;
		}

		for(int i=0;i<X.length;i++){
			if(X[i] == -1){
				X[i]=0;
			}
			ans += Math.pow(10, N-i-1)*X[i];

		}

		if(N>=2 && ans < Math.pow(10, N-1)){
			ans+=Math.pow(10, N-1);
		}

		System.out.println(ans);


	}
}
