Problem Statement

There are $N$ squares arranged in a row from left to right. Let Square $i$ be the $i$-th square from the left.
$M$ of those squares, Square $A_1$, $A_2$, $A_3$, $\\ldots$, $A_M$, are blue; the others are white. ($M$ may be $0$, in which case there is no blue square.)
You will choose a positive integer $k$ just once and make a stamp with width $k$. In one use of a stamp with width $k$, you can choose $k$ consecutive squares from the $N$ squares and repaint them red, as long as those $k$ squares do not contain a blue square.
At least how many uses of the stamp are needed to have no white square, with the optimal choice of $k$ and the usage of the stamp?

Constraints

• $1 \\le N \\le 10^9$
• $0 \\le M \\le 2 \\times 10^5$
• $1 \\le A_i \\le N$
• $A_i$ are pairwise distinct.
• All values in input are integers.

Input

Input is given from Standard Input in the following format:

$N$ $M$ $A_1 \\hspace{7pt} A_2 \\hspace{7pt} A_3 \\hspace{5pt} \\dots \\hspace{5pt} A_M$

Output

Print the minimum number of uses of the stamp needed to have no white square.

Sample Input 1

5 2
1 3

Sample Output 1

3

If we choose $k = 1$ and repaint the three white squares one at a time, three uses are enough, which is optimal.
Choosing $k = 2$ or greater would make it impossible to repaint Square $2$, because of the restriction that does not allow the $k$ squares to contain a blue square.

Sample Input 2

13 3
13 3 9

Sample Output 2

6

One optimal strategy is choosing $k = 2$ and using the stamp as follows:

• Repaint Squares $1, 2$ red.
• Repaint Squares $4, 5$ red.
• Repaint Squares $5, 6$ red.
• Repaint Squares $7, 8$ red.
• Repaint Squares $10, 11$ red.
• Repaint Squares $11, 12$ red.

Note that, although the $k$ consecutive squares chosen when using the stamp cannot contain blue squares, they can contain already red squares.

Sample Input 3

5 5
5 2 1 4 3

Sample Output 3

0

If there is no white square from the beginning, we do not have to use the stamp at all.

Sample Input 4

1 0

Sample Output 4

1

$M$ may be $0$.