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.

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 a single character (quotes for clarity):

• '<' if X < Y
• '>' if X > Y
• '=' if X = Y

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,  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.

#include <iostream>

#include <cstdio>

#include <cstring>

#include <string>

#include <cstdlib>

#include <cmath>

#include <algorithm>

#include <string>

using namespace std;

typedef long long int LL;

int a[50], b[50];

LL pow1(int a, int b) {

	LL res = 1;

	while (b --)

		res *= a;

	return res;

}

int main() {

	memset(a, 0, sizeof(a));

	memset(b, 0, sizeof(b));

	int n, m, k1, k2;

	cin >> n >> k1;

	for (int i = n-1; i >=0; i--)

		cin >> a[i];

	LL res1 = 0;

	for (int i = 0; i < n; i++)

		res1 += a[i] * pow1(k1, i);

	cin >> m >> k2;

	for (int i = m - 1; i >= 0; i--)

		cin >> b[i];

	LL res2 = 0;

	for (int i = 0; i < m; i++)

		res2 += b[i] * pow1(k2, i);

	//cout << res1 << endl << res2 << endl;

	if (res1 == res2)

		cout << "=" << endl;

	else if (res1 > res2) cout << ">" << endl;

	else cout << "<" << endl;

	return 0;

}