【CodeForces 602A】C - 特别水的题3-Two Bases

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=102271#problem/C

Description

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 Input

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

>

Hint

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.

转换进制，一边读入一边转为十进制，再比较大小。

#include<stdio.h>
long long x,y,bx,by,num,decx,decy;
int main(){
scanf("%lld%lld",&x,&bx);
for(int i=0;i<x-1;i++){
scanf("%lld",&num);
decx=(decx+num)*bx;
}
scanf("%lld",&num);
decx=decx+num;
scanf("%lld%lld",&y,&by);
for(int i=0;i<y-1;i++){
scanf("%lld",&num);
decy=(decy+num)*by;
}
scanf("%lld",&num);
decy=decy+num;

if(decx>decy)printf(">\n");
else if(decx<decy)printf("<\n");
else printf("=\n");
return 0;
}