CF1119A.Ilya and a Colorful Walk

Ilya lives in a beautiful city of Chordalsk.

There are $n$ houses on the street Ilya lives, they are numerated from $1$ to $n$ from left to right; the distance between every two neighboring houses is equal to $1$ unit. The neighboring houses are $1$ and $2$, $2$ and $3$, ..., $n-1$ and $n$. The houses $n$ and $1$ are not neighboring.

The houses are colored in colors $c_1, c_2, \ldots, c_n$ so that the $i$-th house is colored in the color $c_i$. Everyone knows that Chordalsk is not boring, so there are at least two houses colored in different colors.

Ilya wants to select two houses $i$ and $j$ so that $1 \leq i \lt j \leq n$, and they have different colors: $c_i \neq c_j$. He will then walk from the house $i$ to the house $j$ the distance of $(j-i)$ units.

Ilya loves long walks, so he wants to choose the houses so that the distance between them is the maximum possible.

Help Ilya, find this maximum possible distance.

伊利亚住在一座美丽的城市——和弦斯克（Chordalsk）。

伊利亚所住的街道上有 $n$ 座房屋，从左到右依次编号为 $1$ 到 $n$；任意两座相邻房屋之间的距离均为 $1$ 单位。相邻的房屋对为：$1$ 与 $2$、$2$ 与 $3$、……、$n-1$ 与 $n$。注意：房屋 $n$ 与 $1$ 不相邻。

这些房屋被涂上了颜色 $c_1, c_2, \ldots, c_n$，即第 $i$ 座房屋的颜色为 $c_i$。众所周知，和弦斯克绝非乏味之地，因此至少有两座房屋颜色不同。

伊利亚希望选出两座房屋 $i$ 和 $j$，满足 $1 \leq i < j \leq n$，且它们颜色不同：$c_i \neq c_j$。接着他将从房屋 $i$ 步行至房屋 $j$，行走距离为 $(j-i)$ 单位。

伊利亚喜爱长途步行，因此他希望所选房屋之间的距离尽可能大。

请帮助伊利亚找出这个可能的最大距离。

The first line contains a single integer $n$ ($3 \leq n \leq 300\,000$) — the number of cities on the street.

The second line contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \leq c_i \leq n$) — the colors of the houses.

It is guaranteed that there is at least one pair of indices $i$ and $j$ so that $1 \leq i \lt j \leq n$ and $c_i \neq c_j$.

第一行包含一个整数 $n$（$3 \leq n \leq 300\,000$）—— 街道上城市的数量。

第二行包含 $n$ 个整数 $c_1, c_2, \ldots, c_n$（$1 \leq c_i \leq n$）—— 各房屋的颜色。

保证至少存在一对下标 $i$ 和 $j$，满足 $1 \leq i \lt j \leq n$ 且 $c_i \neq c_j$。

Print a single integer — the maximum possible distance Ilya can walk.

输出一个整数——伊利亚能行走的最大可能距离。

In the first example the optimal way is to walk from the first house to the last one, where Ilya can walk the distance of $5-1 = 4$ units.

In the second example the optimal way is to either walk from the first house to the second or from the second to the third. Both these ways have the distance of $1$ unit.

In the third example the optimal way is to walk from the third house to the last one, where Ilya can walk the distance of $7-3 = 4$ units.

在第一个例子中，最优方案是从第一栋房子走到最后一栋房子，Ilya 可以行走的距离为 $5-1 = 4$ 个单位。

在第二个例子中，最优方案是 either 从第一栋房子走到第二栋，或从第二栋走到第三栋。这两种方案的行走距离均为 $1$ 个单位。

在第三个例子中，最优方案是从第三栋房子走到最后一栋房子，Ilya 可以行走的距离为 $7-3 = 4$ 个单位。