【42.59%】【codeforces 602A】Two Bases

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard 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 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.

【题目链接】:http://codeforces.com/contest/602/problem/A

【题解】



把它们都转换成10进制再比较就好.

40^10不会爆LL



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int MAXN = 10+5;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

int n,bx,m,by;
LL a[MAXN];

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    rei(n);rei(bx);
    rep2(i,n-1,0)
        rel(a[i]);
    LL x = 0;
    LL now = 1;
    rep1(i,0,n-1)
    {
        x += now*a[i];
        now = now * bx;
    }
    rei(m);rei(by);
    rep2(i,m-1,0)
        rel(a[i]);
    LL y = 0;
    now = 1;
    rep1(i,0,m-1)
    {
        y += now*a[i];
        now = now * by;
    }
    if (x==y)
        putchar('=');
    else
        if (x < y)
            putchar('<');
        else
            putchar('>');
    return 0;
}