释放无限光明的是人心，制造无边黑暗的也是人心，光明和黑暗交织着，厮杀着，这就是我们为之眷恋又万般无奈的人世间。
The string APPAPT contains two PAT's as substrings. The first one is formed by the 2nd, the 4th, and the 6th characters, and the second one is formed by the 3rd, the 4th, and the 6th characters.
Now given any string, you are supposed to tell the number of PAT's contained in the string.
Input Specification:
Each input file contains one test case. For each case, there is only one line giving a string of no more than 10?5?? characters containing only P, A, or T.
Output Specification:
For each test case, print in one line the number of PAT's contained in the string. Since the result may be a huge number, you only have to output the result moded by 1000000007.
Sample Input:
APPAPT
Sample Output:
2#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <vector>
#include <ctime>
#include <cctype>
#include <bitset>
#include <utility>
#include <sstream>
#include <complex>
#include <iomanip>
#include<climits>//INT_MAX
//#include<bits/stdc++.h>
#define PP pair<ll,int>
#define inf 0x3f3f3f3f
#define llinf 0x3f3f3f3f3f3f3f3fll
#define dinf 1000000000000.0
#define PI 3.1415926
typedef long long ll;
using namespace std;
int const mod=1e9+7;
const int maxn=1e6+10;
string s;
int l[maxn],ct;
ll jg;
int main(){getline(cin,s);int cd=s.size();for(int i=0;i<cd;i++){if(s[i]=='P')l[i]=1;if(i!=0)l[i]+=l[i-1];}for(int i=cd-1;i>=0;i--){if(s[i]=='T')ct++;if(s[i]=='A')jg=(jg+l[i]*ct)%mod;}cout<<jg<<endl;return 0;
}