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问题:为了获得工作面试的资格,px必须解决easy函数。作为一名px新手,需要你的帮助才能胜任这份工作。帮助他解决以下简单函数。f(n)是一个由递归关系定义的函数:f(n)=f(n-1)-f(n-2)+n表示n>=2f(0)=1,f(1)=1
Let S(n) be the function defined as :
S(n)= sum(F(i)) i=0 to n. Your task is to calculate the value of S(n) for the given value of n for every test case.
As your answer can be very large print the value of S(n) modulo 1000000007.
限制条件:
1 ≤ T ≤ 100
1 ≤ N ≤ 10^18
输入:
First Line contains T i.e. number of the test case.
Each of the next T lines contains an integer N.
输出:
For each test case print the value of S(N) modulo 1000000007
样本输入:
3
2
5
7
样本输出:
4
21
35
解释
For Test Case:
N=2
F(0) = 1, F(1) = 1
F(2) = F(1) - F(0) + 2
F(2) = 1 - 1 + 2
F(2) = 2
S(2) = F(0) + F(1) + F(2)
S(2) = 1 + 1 + 2
S(2) = 4
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