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: Runtime Error)

var line; var result = 0;
var S; var abc = ["A", "B", "C"];
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("");
    dfs(0, []);
    console.log(result);
});

function dfs(n, comb) {
    if (n === S.length) { calc(comb); return (0); }

    if (S[n] === "?") {
        comb[n] = "A"; dfs(n + 1, comb);
        comb[n] = "B"; dfs(n + 1, comb);
        comb[n] = "C"; dfs(n + 1, comb);
    }
    else {
        comb[n] = S[n]; dfs(n + 1, comb);
    }
}

function calc(comb) {

    var dp = Array(S.length).fill(0)
    for (var i = 0; i < S.length; i++) {
        dp[i] = Array(3).fill(0);
    }

    if (comb[0] === abc[0]) { dp[0][0] += 1; }

    for (var i = 1; i < S.length; i++) {
        if (comb[i] === abc[0]) { dp[i][0] += dp[i - 1][0] + 1; }
        else { dp[i][0] = dp[i - 1][0]; }
    }

    for (var i = 1; i < abc.length; i++) {
        dp[0][i] = 0;
    }

    for (var j = 1; j < 3; j++) {
        var abc_counter = {};
        for (var i = 1; i < S.length; i++) {
            if (abc_counter[comb[i]] === undefined) abc_counter[comb[i]] = 0;

            if (comb[i] === abc[j] && dp[i - 1][j - 1] !== 0) {
                dp[i][j] += dp[i - 1][j - 1] + abc_counter[comb[i]];
                abc_counter[comb[i]]++;
            }
            else {
                dp[i][j] = dp[i - 1][j];
            }
        }
    }
    result += (dp[S.length - 1][2] % MOD);
}