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.*;
import java.util.*;

class Main {

	public static void main(String[] args) throws Exception {
		final Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int M = sc.nextInt();
		int[] a = new int[3];
		Arrays.fill(a, -1);

		for (int i = 0; i < M; i++) {
			int s = sc.nextInt() - 1;
			int c = sc.nextInt();
			if (a[s] != -1 && a[s] != c) {
				System.out.println(-1);
				return;
			}

			if (s + 1 > N) {
				System.out.println(-1);
				return;
			}

			a[s] = c;
		}

		if (N == 1) {
			System.out.println((a[0] != -1) ? a[0] : 0);
			return;
		}

		if (N >= 2 && a[0] == 0) {
			System.out.println(-1);
			return;
		}

		if (a[0] == -1) {
			a[0] = 1;
		}

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

		for (int i = 0; i < N; i++) {
			System.out.print(a[i]);
		}

		System.out.println();
	}
}
