Codeforces Round #471 (Div. 2)B. Not simply beatiful strings

Let's call a string adorable if its letters can be realigned in such a way that they form two consequent groups of equal symbols (note that different groups must contain different symbols). For example, ababa is adorable (you can transform it to aaabb, where the first three letters form a group of a-s and others — a group of b-s), but cccc is not since in each possible consequent partition letters in these two groups coincide.

You're given a string s. Check whether it can be split into two non-empty subsequences such that the strings formed by these subsequences are adorable. Here a subsequence is an arbitrary set of indexes of the string.

Input

The only line contains s (1 ≤ |s| ≤ 105) consisting of lowercase latin letters.

Output

Print «Yes» if the string can be split according to the criteria above or «No» otherwise.

Each letter can be printed in arbitrary case.

Examples

input

ababa

output

Yes

input

zzcxx

output

Yes

input

yeee

output

No

题意：如果能把一个字符串分成两份，每份都可以被分成只有相同字母的两块，并且这两块的字母不相同，那么输出yes，反之no。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
#define eps 0.00000001
#define pn printf("\n")
using namespace std;
typedef long long ll;

const int maxn = 1e5+7;
map <char,int> mp;

int main()
{
    char s[maxn];
    gets(s);
    int len = strlen(s);
    for(int i=0;i<len;i++)
        mp[s[i]]++;
    int fl = 0;
    for(map<char,int>::iterator it = mp.begin(); it!=mp.end(); it++)
        if(it->second == 1)
            fl++;
    if(mp.size() > 4) printf("No\n");
    else if(mp.size() == 2 && fl) printf("No\n");
    else if(mp.size() == 1) printf("No\n");
    else if(mp.size() == fl && fl%2 == 1) printf("No\n");
    else printf("Yes\n");
    
}