cf 602 A(进制转换)

A. Two Bases

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 21 0 1 1 1 12 104 7

output

=

input

3 31 0 22 52 4

output

<

input

7 1615 15 4 0 0 7 107 94 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,  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 <stdio.h>using namespace std;__int64 a[20],b[20];__int64 res,ans;int main(){    int n,m;    res=ans=0;    scanf("%d%d",&n,&m);    for(int i=0;i<n;i++)        scanf("%I64d",a+i);    __int64 aa,bb;    aa=1;    for(int i=n-1;i>=0;i--)    {        res+=a[i]*aa;        aa*=m;    } //   printf("%I64d\n",res);    int x,y;    scanf("%d%d",&x,&y);    for(int i=0;i<x;i++)        scanf("%I64d",b+i);    bb=1;    for(int i=x-1;i>=0;i--)    {        ans+=b[i]*bb;        bb*=y;    } //   printf("%I64d\n",ans);    if(res>ans)        printf(">\n");    else if(res==ans)        printf("=\n");    else        printf("<\n");    return 0;}