Codeforces Round #621 (Div. 1 + Div. 2) C. Cow and Message

Bessie the cow has just intercepted a text that Farmer John sent to Burger Queen! However, Bessie is sure that there is a secret message hidden inside.

The text is a string ss of lowercase Latin letters. She considers a string tt as hidden in string ss if tt exists as a subsequence of ss whose indices form an arithmetic progression. For example, the string aab is hidden in string aaabb because it occurs at indices 11 , 33 , and 55 , which form an arithmetic progression with a common difference of 22 . Bessie thinks that any hidden string that occurs the most times is the secret message. Two occurrences of a subsequence of SS are distinct if the sets of indices are different. Help her find the number of occurrences of the secret message!

For example, in the string aaabb, a is hidden 33 times, b is hidden 22 times, ab is hidden 66 times, aa is hidden 33 times, bb is hidden 11 time, aab is hidden 22 times, aaa is hidden 11 time, abb is hidden 11 time, aaab is hidden 11 time, aabb is hidden 11 time, and aaabb is hidden 11 time. The number of occurrences of the secret message is 66 .

Input

The first line contains a string ss of lowercase Latin letters (1≤|s|≤1051≤|s|≤105 ) — the text that Bessie intercepted.

Output

Output a single integer  — the number of occurrences of the secret message.

Examples

Input

aaabb

Output

6

Input

usaco

Output

1

Input

lol

Output

2

Note

In the first example, these are all the hidden strings and their indice sets:

• a occurs at (1)(1) , (2)(2) , (3)(3)
• b occurs at (4)(4) , (5)(5)
• ab occurs at (1,4)(1,4) , (1,5)(1,5) , (2,4)(2,4) , (2,5)(2,5) , (3,4)(3,4) , (3,5)(3,5)
• aa occurs at (1,2)(1,2) , (1,3)(1,3) , (2,3)(2,3)
• bb occurs at (4,5)(4,5)
• aab occurs at (1,3,5)(1,3,5) , (2,3,4)(2,3,4)
• aaa occurs at (1,2,3)(1,2,3)
• abb occurs at (3,4,5)(3,4,5)
• aaab occurs at (1,2,3,4)(1,2,3,4)
• aabb occurs at (2,3,4,5)(2,3,4,5)
• aaabb occurs at (1,2,3,4,5)(1,2,3,4,5)

Note that all the sets of indices are arithmetic progressions.

In the second example, no hidden string occurs more than once.

In the third example, the hidden string is the letter l.

大意是给定一个字符串，让你找出所有下标成等差数列的子串里出现次数最多的出现了多少次。

看了别人的博客才勉强搞懂-_- 首先要说的就是一个字母也算是n=1的等差数列。然后我们可以发现，n=1，2，3...当n>=3以后，形成等差数列的条件越强，一个n=3的等差数列可以拆分成三个n等于2的等差数列，因此我们可以直接忽略掉n>=3时的情况。n=1时，直接开一个数组single[26]统计每个字母出现的次数。n=2时会麻烦一点，这里使用一个二维数组cnt[26][26]统计，其含义为：cnt[i][j]代表字母对('a'+i, 'a'+j)出现的次数，一定注意顺序！然后扫一遍字符串，第一维获取当前字母，然后再写一个0~26的二重循环，更新a~z与当前字母组成的字母对的个数，注意当前字母是第二个。更新的核心操作是cnt[j][s[i]-'a']+=single[j] （i是第一维循环变量，j是第二维循环变量),直接加上j对应的字母之前出现的次数即可。

别忘开long long！！

#include <bits/stdc++.h>
using namespace std;
string s;
long long cnt[26][26]={0};
int single[26]={0};
long long mmax(long long a, long long b)
{
    return a>b?a:b;
}
int main()
{
    cin>>s;
    int i,j;
    if(s.size()==1)
    {
        cout<<1;
        return 0;
    }
    long long ans=0;
    for(i=0;i<s.size();i++)
    {
        for(j=0;j<26;j++)
        {
            cnt[j][s[i]-'a']+=single[j];
            ans=mmax(ans,cnt[j][s[i]-'a']);
        }
        single[s[i]-'a']++;//更新n=1情况时一定要放在后面，要不然的话会被第二重循环统计到当前的字母 
        ans=mmax(ans,single[s[i]-'a']);
    }
    cout<<ans;
    return 0;
}