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 JavaScript (node.js v5.12) (judge verdict: Accepted)

var line; var S;
var MOD = Math.pow(10, 9) + 7;

var readline = require('readline');
var rl = readline.createInterface({
    input: process.stdin,
    output: process.stdout,
});

rl.on('line', function (x) {
    line = x;
});

rl.on('close', function () {
    S = line.split("");
    dp = Array(S.length + 1);
    
    for (var i = 0; i <= S.length; i++) {
        dp[i] = Array(3 + 1).fill(0);
    }

    for (var i = S.length; i >= 0; i--) {
        for (var j = 3; j >= 0; j--) {
            if (i === S.length) { dp[i][j] = (j === 3 ? 1 : 0); }
            else {
                dp[i][j] = dp[i + 1][j] * (S[i] === '?' ? 3 : 1);
                if (j < 3 && (S[i] === '?' || S[i] === "ABC"[j])) {
                    dp[i][j] += dp[i + 1][j + 1];
                }
                dp[i][j] %= MOD;
            }
        }
    }
    console.log(dp[0][0]);
});