A scientist was doing a research on some kinds of bacteria. He found that the kinds, he examined, take T unit of time to grow (be mature) enough in order to can reproduce.

Also he found that each type reproduces with a constant rate which is N new bacteria every F unit of time.

(where F=T)

Task

write a progam that reads L (number of bacteria (at the begining of the experiment)), M (number of mature bacteria of them), T (time of each to get mature which is also the time needed for reproducing N new bacteria), N (rate of reproducing per T unit of time) and Z (period elapsed by the experiment).

Calculate the number of bacteria after Z unit of time.Regardless of life-span

Constraints

1 ≤ L ≤ 5     number of bacteria (at the begining of the experiment)

1 ≤ M ≤ L    number of mature bacteria

1 ≤ T ≤ 5   time of each to get mature which is also the time needed for reproducing N new bacteria

1 ≤ N ≤ 50    rate of reproducing per T unit of time

1 ≤ Z/T ≤ 4,300    period elapsed by the experiment

Note

Z is always divisible by T.

Input

• L (number of bacteria (at the begining of the experiment))
• M (number of mature bacteria of them)
• T (time of each to get mature which is also the time needed for reproducing N new bacteria)
• N (rate of reproducing per T unit of time)
• Z (period elapsed by the experiment)

Output

• the number of bacteria after Z unit of time.Regardless of life-span.

Example

Input:
3
2
3
1
3

Output:
5

The experiment begins with 2 mature bacteria and one unmature bacterium.
For, each of the mature bacteria reproduces after 3 units of time.
Then th total becomes 4 -as each one got a new one (2*2)-.
But, for the unmature bacterium after 3 units of time, it only become mature.
After all of that the experiment finishes with 5 bacteria.


Input
2
0
1
1
100

Output:
1146295688027634168202