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 program that reads L (number of bacteria (at the beginning 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 beginning of the experiment).
• 0 ≤ 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 beginning 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

Input
2
0
1
1
100

Output:
1146295688027634168202

Explanation for Test Case 1

The experiment begins with 2 mature bacteria and one immature 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 immature bacterium after 3 units of time, it only become mature. After all of that the experiment finishes with 5 bacteria.