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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

#define int long long 
#define INF 1e9
#define LINF 1e18

typedef unsigned long long ull;
typedef pair<int, int> P;
typedef pair<int, P> Pi;
typedef pair<P, P> PP;

const int MOD = 1e9 + 7;
const int dy[] = { 0, 0, 1, -1 };
const int dx[] = { 1, -1, 0, 0 };

template <class T> void chmin(T& a, const T& b) { a = min(a, b); }
template <class T> void chmax(T& a, const T& b) { a = max(a, b); }

int dp[100001][4];

signed main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    string s;
    cin >> s;

    dp[0][0] = 1;
    for(int i = 0; i < s.size(); ++i) {
        for(int j = 0; j <= 3; ++j) {
            if(s[i] == '?') dp[i + 1][j] = (dp[i + 1][j] + (dp[i][j] * 3) % MOD) % MOD;
            else dp[i + 1][j] = (dp[i + 1][j] + dp[i][j]) % MOD;
        }
        for(int j = 0; j < 3; ++j) {
            if(s[i] == "ABC"[j] || s[i] == '?') {
                dp[i + 1][j + 1] = dp[i + 1][j + 1] + dp[i][j] % MOD;
            }
        }
    }
    cout << dp[s.size()][3] << endl;
    return 0;
}