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The electrical resistance is the opposition to the passage of electric current. If two resistors with resistance R1 and R2 are connected to each other, the resultant resistance R depends on how their ends are connected. If they are connected in Series, they simply add up to give R = R1 + R2. If they are connected in Parallel, its given by the equation 1/R = 1/R1 + 1/R2. This is shown below.
We have a long circuit of resistors of one unit resistance each ( Ri = 1 ) having N blocks. Each block has 2 resistors as shown in the figure below. For a circuit with only one block (N=1), the equivalent resistance at the ends on left-side is R = 1 ( note that the right-most horizontally aligned resistor has no influence, as its not part of any simple path between the two left ends ). For a circuit with two blocks (N=2), its 2/3 units. This is shown in the figure below.
Given N, find the resistance A/B at the left ends of the circuit having N blocks. The values can be huge, so reduce the fraction A/B to lowest terms P/Q, where greatest_common_divisor_of (P,Q) = 1 and print (P%M)/(Q%M) for a given modulo M.
First line contains T ( number of test cases, 1 ≤ T ≤ 10 ). Each of the next T lines contains N M ( 1 <= N <= 106 and 2 <=M <= 230 )
For each test case, output the equivalent resistance at the left ends, in the form (P%M)/(Q%M) as explained above, in a separate line.
Input: 3 1 10 2 12 100 10000 Output: 1/1 2/3 4301/9525
|Tags||cook21, easy, flying_ant|
|Time Limit:||5 sec|
|Source Limit:||50000 Bytes|
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