Problem:
Score : 400 points

Problem Statement

The ABC number of a string T is the number of triples of integers (i, j, k) that satisfy all of the following conditions:

1 ≤ i < j < k ≤ |T| (|T| is the length of T.)

T_i = A (T_i is the i-th character of T from the beginning.)

T_j = B

T_k = C

For example, when T = ABCBC, there are three triples of integers (i, j, k) that satisfy the conditions: (1, 2, 3), (1, 2, 5), (1, 4, 5). Thus, the ABC number of T is 3.

You are given a string S. Each character of S is A, B, C or ?.

Let Q be the number of occurrences of ? in S. We can make 3^Q strings by replacing each occurrence of ? in S with A, B or C. Find the sum of the ABC numbers of all these strings.

This sum can be extremely large, so print the sum modulo 10^9 + 7.

Constraints

3 ≤ |S| ≤ 10^5

Each character of S is A, B, C or ?.

Input

Input is given from Standard Input in the following format:

S

Output

Print the sum of the ABC numbers of all the 3^Q strings, modulo 10^9 + 7.

Sample Input 1

A??C

Sample Output 1

8

In this case, Q = 2, and we can make 3^Q = 9 strings by by replacing each occurrence of ? with A, B or C. The ABC number of each of these strings is as follows:

AAAC: 0

AABC: 2

AACC: 0

ABAC: 1

ABBC: 2

ABCC: 2

ACAC: 0

ACBC: 1

ACCC: 0

The sum of these is 0 + 2 + 0 + 1 + 2 + 2 + 0 + 1 + 0 = 8, so we print 8 modulo 10^9 + 7, that is, 8.

Sample Input 2

ABCBC

Sample Output 2

3

When Q = 0, we print the ABC number of S itself, modulo 10^9 + 7. This string is the same as the one given as an example in the problem statement, and its ABC number is 3.

Sample Input 3

????C?????B??????A???????

Sample Output 3

979596887

In this case, the sum of the ABC numbers of all the 3^Q strings is 2291979612924, and we should print this number modulo 10^9 + 7, that is, 979596887.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <list>
#include <queue>
#include <deque>
#include <algorithm>
#include <numeric>
#include <utility>
#include <complex>
#include <functional>
using namespace std;
const int MOD = 1000000007;
typedef long long ll;
typedef pair<ll, ll> p;
const int INF = (1 << 28);
const int dx[4] = {0, 1, 0, -1}, dy[4] = {-1, 0, 1, 0};
const int Dx[8] = {0, 1, 1, 1, 0, -1, -1, -1}, Dy[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
#define yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define no cout << "No" << endl
#define NO cout << "NO" << endl
#define REP(i, n) for (int i = 0; i < n; i++)
#define REPR(i, n) for (int i = n; i >= 0; i--)
#define FOR(i, m, n) for (int i = m; i < n; i++)
#define INF 2e9
#define ALL(v) v.begin(), v.end()
string s;
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> s;
    ll count[3];
    count[0]=0;
    count[1]=0;
    count[2]=1;
    ll ans=0;
    for(int i=0;i<s.size();i++){
        if(s[i]=='A'){
            count[0]+=count[2];
        }
        if(s[i]=='B'){
            count[1]+=count[0];
            count[1]%=MOD;
        }
        if(s[i]=='C'){
            ans+=count[1];
            ans%=MOD;
        }
        if(s[i]=='?'){
            ll temp1=count[0], temp2=count[1], temp3=ans, temp4=count[2];
            count[0]+=temp4;
            count[0]+=temp1*2;
            count[0]%=MOD;
            count[1]+=temp1;
            count[1]+=temp2*2;
            count[1]%=MOD;
            ans+=temp2;
            ans%=MOD;
            ans+=temp3*2;
            ans%=MOD;
            count[2]*=3;
            count[2]%=MOD;
        }
    }
    cout << ans << "\n";
}