Two Bases

Two Bases

 CodeForces - 602A 

After seeing the "ALL YOUR BASE ARE BELONG TO US" meme for the first time, numbers Xand 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 mspace-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 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,  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.

题目大意：输入数字的位数以及是几进制的数，比较两个数的大小

AC代码

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <set>
#include <utility>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define lep(i,l,r) for(int i=l;i>=r;i--)
#define ms(arr) memset(arr,0,sizeof(arr))
//priority_queue<int,vector<int> ,greater<int> >q;
const int maxn = (int)1e5 + 5;
const ll mod = 1e9+7;
ll shi(int n,int b,int arr[])
{
	ll ans=1,base=1;
	for(int i=n;i>=1;i--) {
		ans=ans+base*arr[i];
		base=base*b;
	}
	return ans;
}
int main() 
{
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    ios::sync_with_stdio(0),cin.tie(0);
    int n,b,arr[20];
    cin>>n>>b;
    rep(i,1,n) {
    	cin>>arr[i];
    }
    ll ans1=shi(n,b,arr);
    cin>>n>>b;
    rep(i,1,n) {
    	cin>>arr[i];
    }
    ll ans2=shi(n,b,arr);
    if(ans1==ans2)
    	cout<<"="<<endl;
    else if(ans1<ans2)
    	cout<<"<"<<endl;
    else
    	cout<<">"<<endl;
    return 0;
}