602_A. Two Bases
After seeing the "ALL YOUR BASE ARE BELONG TO US" meme for the first time, numbers X and Y realised that they have different bases, which complicated their relations.

You're given a number X represented in base bx and a number Y represented in base by. Compare those two numbers.

Input

The first line of the input contains two space-separated integers n and bx (1 ≤ n ≤ 10, 2 ≤ bx ≤ 40), where n is the number of digits in the bx-based representation of X. 

The second line contains n space-separated integers x1, x2, ..., xn (0 ≤ xi < bx) — the digits of X. They are given in the order from the most significant digit to the least significant one.

The following two lines describe Y in the same way: the third line contains two space-separated integers m and by (1 ≤ m ≤ 10, 2 ≤ by ≤ 40, bx ≠ by), where m is the number of digits in the by-based representation of Y, and the fourth line contains m space-separated integers y1, y2, ..., ym (0 ≤ yi < by) — the digits of Y.

There will be no leading zeroes. Both X and Y will be positive. All digits of both numbers are given in the standard decimal numeral system.

Output

Output a single character (quotes for clarity): 

  * '<' if X < Y
  * '>' if X > Y
  * '=' if X = Y

Examples

Input

6 2
1 0 1 1 1 1
2 10
4 7


Output

=


Input

3 3
1 0 2
2 5
2 4


Output

&lt;


Input

7 16
15 15 4 0 0 7 10
7 9
4 8 0 3 1 5 0


Output

&gt;

Note

In the first sample, X = 1011112 = 4710 = Y.

In the second sample, X = 1023 = 215 and Y = 245 = 1123, thus X < Y.

In the third sample, <image> and Y = 48031509. We may notice that X starts with much larger digits and bx is much larger than by, so X is clearly larger than Y.
```python
# print ("Input digits and base for number one")
n1, b1 = (int(x) for x in input().split())
# print ("Input all the digits")
d1 = list(int(x) for x in input().split())
d1.reverse()

# print ("Input digits and base for number two")
n2, b2 = (int(x) for x in input().split())
# print ("Input all the digits")
d2 = list(int(x) for x in input().split())
d2.reverse()

# Compute base ten representation of number one
answer1 = 0
power1 = 1
for digit in d1:
    answer1 += digit*power1
    power1 *= b1
# print(answer1)

# Compute base ten representation of number two
answer2 = 0
power2 = 1
for digit in d2:
    answer2 += digit*power2
    power2 *= b2
# print(answer2)

if answer1 < answer2:
    print("<")
elif answer2 < answer1:
    print(">")
else:
    print("=")

    


```