CF209A.Multicolored Marbles

Polycarpus plays with red and blue marbles. He put n marbles from the left to the right in a row. As it turned out, the marbles form a zebroid.

A non-empty sequence of red and blue marbles is a zebroid, if the colors of the marbles in this sequence alternate. For example, sequences (red; blue; red) and (blue) are zebroids and sequence (red; red) is not a zebroid.

Now Polycarpus wonders, how many ways there are to pick a zebroid subsequence from this sequence. Help him solve the problem, find the number of ways modulo 1000000007 (109 + 7).

波利卡普斯玩红色和蓝色的弹珠。他从左到右将 $ n $ 颗弹珠排成一行。结果发现，这些弹珠构成了一条“斑马条纹序列”（zebroid）。

一个非空的红蓝弹珠序列被称为斑马条纹序列（zebroid），当且仅当该序列中弹珠的颜色严格交替出现。例如，序列（红；蓝；红）和（蓝）是斑马条纹序列，而序列（红；红）则不是。

现在波利卡普斯想知道：从这个序列中选出一个斑马条纹子序列，共有多少种方法？请你帮他解决这个问题，求出方案数对 $ 1000000007 $（即 $ 10^9 + 7 $）取模的结果。

The first line contains a single integer n (1 ≤ n ≤ 106) — the number of marbles in Polycarpus's sequence.

第一行包含一个整数 $n$（$1 \leq n \leq 10^6$）—— Polycarpus 序列中弹珠的数量。

Print a single number — the answer to the problem modulo 1000000007 (109 + 7).

输出一个数字——该问题答案对 1000000007（$10^9 + 7$）取模的结果。

Let's consider the first test sample. Let's assume that Polycarpus initially had sequence (red; blue; red), so there are six ways to pick a zebroid:

• pick the first marble;
• pick the second marble;
• pick the third marble;
• pick the first and second marbles;
• pick the second and third marbles;
• pick the first, second and third marbles.

It can be proven that if Polycarpus picks (blue; red; blue) as the initial sequence, the number of ways won't change.

我们来考虑第一个测试样例。假设 Polycarpus 最初拥有序列（红色；蓝色；红色），那么一共有六种方式选出一个斑马序列（zebroid）：

• 选取第一颗弹珠；
• 选取第二颗弹珠；
• 选取第三颗弹珠；
• 选取第一颗和第二颗弹珠；
• 选取第二颗和第三颗弹珠；
• 选取第一颗、第二颗和第三颗弹珠。

可以证明，若 Polycarpus 选择（蓝色；红色；蓝色）作为初始序列，则选取方式的总数不会改变。