spoj#MAIN74. Euclids algorithm revisited
Euclids algorithm revisited
Consider the famous euclid algoithm to calculate the GCD of two integers (a, b):
int gcd(int a, int b) { while (b != 0) { int temp = a; a = b; b = temp % b; } return a; }
for input (7, 3) the 'while' loop will run 2 times as follows: (7, 3) => (3, 1) => (1, 0)
Now given an integer N you have to find the smallest possible sum of two non-negative integers a, b (a >= b) such that the while loop in the above mentioned function for (a, b) will run exactly N times.
Input
First line of input contains T (1 <= T <= 50) the number of test cases. Each of the following T lines contains an integer N (0 <= N <= 10^18).
Output
For each test case print the required answer modulo 1000000007 in a seperate line.
Example
Input: 1 1</p>Output: 2
Explaination: (1,1) is the required pair.