PROBLEM LINK:
Author & Editorialist: Karthikeya
DIFFICULTY:
Easy
PREREQUISITES:
None
PROBLEM:
Given an integer N, find the number of distinct N bit binary numbers such that no two consecutive bits are 1 in each of them.
QUICK EXPLANATION:
By deciding on whether the least significant bit is 0 or 1 for each arrangement, we can see that:
Count(N) = Count(N1)+Count(N2)
So this is a Fibonacci series with the first two terms being 2 and 3.
EXPLANATION:
Let us denote the number of distinct N bit numbers that do not have two consecutive 1's by Count(N).
So as we can see in the figure, Count(1) is 2 and Count(2) is 3.
Now to get Count(3) which is for N=3, let us try to add 1 bit at the end of all binary numbers that we have for N=2.

We can add 0 at the end to 00, 01 and 10 and we get 000,010 and 100 respectively. As we added 0 at the end all the new numbers satisfy the condition that there should not be consecutive 1's. So from here we get Count(2) distinct numbers for N=3 having 0 as the least significant bit.  (1)

In similar way we cannot add 1 at the end of all the distinct numbers of N=2 as we would get consecutive 1's in few cases. So to add a 1 at the end, the last but one bit should be 0. So we can count these numbers by adding 01 at the end of all distinct numbers of N=2. So from here we get Count(1) disticnt numbers for N=3 having 1 as the least significant bit.  (2)
So from 1 and 2: Count(3) = Count(1) + Count(2).
Similarily for every N we get Count(N) = Count(N2) + Count(N1).
So this is a Fibonacci series with the first two terms being 2 and 3.
Hence we precompute the first N numbers of the Fibonacci series with modulus 10^9+7 as A[i] = (A[i1]+A[i2])\% 10^9+7 where A[1]=2 and A[2]=3 (A is an array of size 10^6). Then answer each query in O(1).
TIME COMPLEXITY:
Time: O(N+T)
Space: O(N)
SOLUTION:
Setter's Solution
Setters Solution:
M = 10**9+7
# initializing array before precomputing
Count = [i+1 for i in range(10**6+2)]
# Precomputation
for i in range(2,10**6+2):
Count[i]= (Count[i1]+Count[i2])%M
# Answering Queries
for _ in range(int(input())):
print(l[int(input())])