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

Sample test(s)

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
<

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

output
>

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, $\text{[math]}$ 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.

此题坑点“pow函数会损失精度！

#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;
}