CF 353D

题意： if at some time the i-th position has a boy and the (i + 1)-th position has a girl, then in a second, the i-th position will have a girl and the (i + 1)-th one will have a boy

求什么时候不在交换（时间）

D. Queue

There are n schoolchildren, boys and girls, lined up in the school canteen in front of the bun stall. The buns aren't ready yet and the line is undergoing some changes.

Each second all boys that stand right in front of girls, simultaneously swap places with the girls (so that the girls could go closer to the beginning of the line). In other words, if at some time the i-th position has a boy and the (i + 1)-th position has a girl, then in a second, the i-th position will have a girl and the (i + 1)-th one will have a boy.

Let's take an example of a line of four people: a boy, a boy, a girl, a girl (from the beginning to the end of the line). Next second the line will look like that: a boy, a girl, a boy, a girl. Next second it will be a girl, a boy, a girl, a boy. Next second it will be a girl, a girl, a boy, a boy. The line won't change any more.

Your task is: given the arrangement of the children in the line to determine the time needed to move all girls in front of boys (in the example above it takes 3 seconds). Baking buns takes a lot of time, so no one leaves the line until the line stops changing.

Input

The first line contains a sequence of letters without spaces s1s2... sn (1 ≤ n ≤ 106), consisting of capital English letters M and F. If letter si equals M, that means that initially, the line had a boy on the i-th position. If letter siequals F, then initially the line had a girl on the i-th position.

Output

Print a single integer — the number of seconds needed to move all the girls in the line in front of the boys. If the line has only boys or only girls, print 0.

Sample test(s)

input

MFM

output

1

input

MMFF

output

3

input

FFMMM

output

0

Note

In the first test case the sequence of changes looks as follows: MFM  →  FMM.

The second test sample corresponds to the sample from the statement. The sequence of changes is: MMFF  → MFMF  →  FMFM  →  FFMM.

//#pragma comment(linker, "/STACK:102400000")
#include<cstdlib>
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
#include<list>
#include<queue>
#include<vector>
#define tree int o,int l,int r
#define lson o<<1,l,mid
#define rson o<<1|1,mid+1,r
#define lo o<<1
#define ro o<<1|1
#define pb push_back
#define mp make_pair
#define ULL unsigned long long
#define LL long long
#define inf 0x7fffffff
#define eps 1e-7
#define N 1000009
using namespace std;
int m,n,T,t,x,y,u;
char str[N];
int main()
{
#ifndef ONLINE_JUDGE
    freopen("ex.in","r",stdin);
#endif
    while(scanf("%s",str)==1)
    {
        int ans=0,cnt=0;
        n=strlen(str);
        for(int i=0;i<n;i++)
        {
            if(str[i]=='M')cnt++;
            else if(cnt!=0)
            {
                ans=max(ans+1,cnt);
            }
        }
        printf("%d
",ans);

    }
    return 0;
}