Unlucky year in Berland is such a year that its number n can be represented as n = x^a + y^b, where a and b are non-negative integer numbers.

For example, if x = 2 and y = 3 then the years 4 and 17 are unlucky (4 = 2⁰ + 3¹

Such interval of years that there are no unlucky years in it is called The Golden Age.

You should write a program which will find maximum length of The Golden Age which starts no earlier than the year l

Input

The first line contains four integer numbers x, y, l and r (2 ≤ x, y ≤ 10¹⁸, 1 ≤ l ≤ r ≤ 10¹⁸

Output

Print the maximum length of The Golden Age within the interval [l, r].

If all years in the interval [l, r] are unlucky then print 0.

Examples

2 3 1 10

1

3 5 10 22

8

2 3 3 5

0

Note

In the first example the unlucky years are 2, 3, 4, 5, 7, 9 and 10. So maximum length of The Golden Age is achived in the intervals [1, 1], [6, 6] and [8, 8].

In the second example the longest Golden Age is the interval [15, 22].

思路：直接枚舉a和b的值，因為2^60大於1E18，所以可以0~61的任意枚舉。

把可以得出的數字加入到一個set中，最後從頭到尾遍歷找最大的區間。

註意：由於數字很大，稍微大一點就爆了longlong 所以判斷是否大於R的時候，用自帶函數pow，因為float/double的範圍很大，比LL大多了，所以不會炸精度，可以剔除掉過大的數據，

而加入到set的時候用快速冪來算那個數值，因為浮點數有精度誤差。還讀到了大佬的博客是把除法改成乘法來防止爆longlong，受益匪淺。

推薦一個：https://blog.csdn.net/qq_37129433/article/details/81667733

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), ‘\0‘, sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define gg(x) getInt(&x)
using namespace std;
typedef  long long ll;
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
ll x,y,l,r;
vector<ll> vc;
set<ll> v;
ll quick_pow(ll a,ll b)
{
    ll ans=1;
        while(b)
        {
                if(b&1) ans= ( ans * a);
                a=( a* a);
                b>>=1;
        }
        return ans;
}
int main()
{
    gbtb;
    cin>>x>>y>>l>>r;
    for(ll i=0ll;i<=63ll;i++)
    {
        if(pow(x,i)>r||pow(x,i)<=0)
        {
            break;
        }
        for(ll j=0ll;j<=63ll;j++)
        {
            long long k=1ll*quick_pow(x,i)+1ll*quick_pow(y,j);
//            if(quick_pow(x,i)>r||quick_pow(x,i)<=0)
//            {
//                break;
//            }
            if(pow(y,j)>r||pow(y,j)<=0)
            {
                break;
            }
            if(k<=0||k>r)
            {
                break;
            }else
            {
                if(k>=l&&k<=r)
                {
                    v.insert(k);
                }
            }
        }
    }
    set<ll> ::iterator it=v.begin();
    ll ans=0ll;
    ll last=l-1;
    ll now;
    while(it!=v.end())
    {
//        cout<<*it<<" ";
         now=*it;
        if(now-last-1>ans)
        {
            ans=now-last-1;
        }
        last=now;
        it++;
    }
    r-now;
    now=r+1;
    if(now-last-1>ans)
    {
        ans=now-last-1;
    }
    cout<<ans<<endl;
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ‘ ‘ || ch == ‘\n‘);
    if (ch == ‘-‘) {
        *p = -(getchar() - ‘0‘);
        while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) {
            *p = *p * 10 - ch + ‘0‘;
        }
    }
    else {
        *p = ch - ‘0‘;
        while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) {
            *p = *p * 10 + ch - ‘0‘;
        }
    }
}

The Golden Age CodeForces - 813B （數學+枚舉）