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 b_x and a number Y represented in base b_y. Compare those two numbers.

Input

The first line of the input contains two space-separated integers n and b_x (1 ≤ n ≤ 10, 2 ≤ b_x ≤ 40), where n is the number of digits in the b_x-based representation of X.

The second line contains n space-separated integers x₁, x₂, ..., x_n (0 ≤ x_i < b_x) — 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 b_y (1 ≤ m ≤ 10, 2 ≤ b_y ≤ 40, b_x ≠ b_y), where m is the number of digits in the b_y-based representation of Y, and the fourth line contains m space-separated integers y₁, y₂, ..., y_m (0 ≤ y_i < b_y) — 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 b_x is much larger than b_y, so X is clearly larger than Y.

 1 #include<cstdio>
 2 #include<vector>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<algorithm>
 6 using namespace std;
 7 typedef long long ll;
 8 typedef unsigned long long ull;
 9 int main()
10 {
11     ull a,b,bx,by,res1=0,res2=0;
12     int x,y,m,n;
13     scanf("%d%I64d",&n,&bx);
14     for(int i=0;i<n;i++)
15     {
16         scanf("%I64d",&a);
17         res1+=a*pow(bx*1.0,n-i-1);
18     }
19     scanf("%d%I64d",&m,&by);
20     for(int i=0;i<m;i++)
21     {
22         scanf("%I64d",&b);
23         res2+=b*pow(by*1.0,m-i-1);
24     }
25     if(res1>res2)
26         puts(">");
27     else if(res1==res2)
28         puts("=");
29     else
30         puts("<");
31     return 0;
32 }