Educational Codeforces Round 22.B 暴力

B. The Golden Age

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Unlucky year in Berland is such a year that its number n can be represented as n = xa + yb, 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, 17 = 23 + 32 = 24 + 30) and year 18 isn't unlucky as there is no such representation for it.

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 and ends no later than the year r. If all years in the interval [l, r] are unlucky then the answer is 0.

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

input

2 3 1 10

output

1

input

3 5 10 22

output

8

input

2 3 3 5

output

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].

题意：给定一个a和b 如果满足 pow(a,i)+pow(b,j),就说这个数不幸运 求一个区间内最大的没有不幸运数的区间。

思路：枚举a^i 与 b^j的值，存入vector, 计算结果

代码：

 #include<stdio.h>
 #include<iostream>
 #include<algorithm>
 #include<string.h>
 #include<math.h>
 #include<stdlib.h>
 #include<ctype.h>
 #include<stack>
 #include<queue>
 #include<map>
 #include<set>
 #include<vector>
 #define ll long long
 #define  db double
 using namespace std;
 ;
 ;
 vector<ll>a;
 ll x,y,l,r,ans=;
 int main(){
     scanf("%lld%lld%lld%lld",&x,&y,&l,&r);
     a.push_back(l-);
     a.push_back(r+);
     ;;i*=x){
         ;;j*=y){
             ll len=i+j;
             if(len>=l&&len<=r)a.push_back(i+j);
             if(j>r/y)break;//判断条件必须单写，若改为i<=r,j<=r 会MLE，因为x，y很大
         }
         if(i>r/x)break;
     }
     sort(a.begin(),a.end());
     ;i<a.size();i++) ans=max(ans,a[i]-a[i-]-);
     printf("%lld\n",ans);
 }