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
#include <stdio.h>
#include <math.h>

int s[11], c[11], d[11], n;

void digit(int N)
{
	int cnt = n;

	while (N)
	{
		d[cnt--] = N % 10;
		N /= 10;
	}
}

int main(void)
{
	int m, i, j;

	scanf("%d %d", &n, &m);
	for (i = 1; i <= m; i++)
		scanf("%d %d", &s[i], &c[i]);

	for (i = n > 1 ? (int)pow(10.0, n - 1.0) : 0; i < (int)pow(10.0, n); i++)
	{
		digit(i);

		for (j = 1; j <= m; j++)
			if (d[s[j]] != c[j])
				break;
		if (j == m + 1)
		{
			printf("%d", i);
			return 0;
		}
	}
	puts("-1");
	return 0;
}