Codeforces Round #333 (Div. 2)

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

Examples

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

>

注意范围

#include<stdio.h>
#include<math.h>
#define LL __int64
int main()
{
int n,m;
    int u,v,a,b;
double x,y,ans;
while(~scanf("%d%d",&n,&a))
{
        ans=pow(a,n-1);
        x=0;
        while(n--)
        {
        scanf("%d",&u);
        x+=u*ans;
        ans/=a;
}
y=0;
scanf("%d%d",&m,&b);
ans=pow(b,m-1);
while(m--)
{
scanf("%d",&v);
y+=v*ans;
ans/=b;
}
if(x==y)
printf("=\n");
else if(x>y)
printf(">\n");
else
printf("<\n");
}
return 0;
}