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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

const double PI = 3.14159265358979323846;
typedef long long ll;
const double EPS = 1e-9;
#define rep(i, n) for (int i = 0; i < (n); ++i)
//#define rep(i, n) for (ll i = 0; i < (n); ++i)
//typedef pair<ll, ll> P;
typedef pair<double, double> P;
const ll INF = 10e17;
#define cmin(x, y) x = min(x, y)
#define cmax(x, y) x = max(x, y)
#define ret() return 0;

double equal(double a, double b) {
    return fabs(a - b) < DBL_EPSILON;
}

std::istream &operator>>(std::istream &in, set<string> &o) {
    string a;
    in >> a;
    o.insert(a);
    return in;
}

std::istream &operator>>(std::istream &in, queue<int> &o) {
    ll a;
    in >> a;
    o.push(a);
    return in;
}

bool contain(set<int> &s, int a) { return s.find(a) != s.end(); }

//ofstream outfile("log.txt");
//outfile << setw(6) << setfill('0') << prefecture << setw(6) << setfill('0') << rank << endl;
// std::cout << std::bitset<8>(9);
//const ll mod = 1e10;

typedef priority_queue<ll, vector<ll>, greater<ll> > PQ_ASK;
const int mod = 1000000007;

struct mint {
    ll x; // typedef long long ll;
    mint(ll x = 0) : x((x % mod + mod) % mod) {}

    mint &operator+=(const mint a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }

    mint &operator-=(const mint a) {
        if ((x += mod - a.x) >= mod) x -= mod;
        return *this;
    }

    mint &operator*=(const mint a) {
        (x *= a.x) %= mod;
        return *this;
    }

    mint operator+(const mint a) const {
        mint res(*this);
        return res += a;
    }

    mint operator-(const mint a) const {
        mint res(*this);
        return res -= a;
    }

    mint operator*(const mint a) const {
        mint res(*this);
        return res *= a;
    }

    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t >> 1);
        a *= a;
        if (t & 1) a *= *this;
        return a;
    }

    // for prime mod
    mint inv() const {
        return pow(mod - 2);
    }

    mint &operator/=(const mint a) {
        return (*this) *= a.inv();
    }

    mint operator/(const mint a) const {
        mint res(*this);
        return res /= a;
    }

    friend std::istream &operator>>(std::istream &in, mint &o) {
        ll a;
        in >> a;
        o = a;
        return in;
    }

    friend std::ostream &operator<<(std::ostream &out, const mint &o) {
        out << o.x;
        return out;
    }

};

class CumulativeSum {
    vector<ll> numbers;
    vector<ll> sums;

public:
    CumulativeSum(int n) {
        numbers.resize(n);
        sums.resize(n);
    }

    void set(int i, ll value) {
        numbers[i] = value;
    }

    ll getSum(int i) {
        if (i == -1) return 0;
        if (i == sums.size()) return sums.back();
        return sums[i];
    }

    ll getSectionSum(int start, int end) {
        return getSum(end) - getSum(start - 1);
    }

    void calculate() {
        for (int i = 0; i < numbers.size(); i++) {
            sums[i] = getSum(i - 1) + numbers[i];
        }
    }

};


int main() {

    string s;
    cin >> s;
    int n = s.size();

    CumulativeSum as(n), bs(n), cs(n), hs(n);
    rep(i, n) {
        char c = s[i];
        if (c == 'A') as.set(i, 1);
        if (c == 'B') bs.set(i, 1);
        if (c == 'C') cs.set(i, 1);
        if (c == '?') hs.set(i, 1);
    }
    as.calculate();
    bs.calculate();
    cs.calculate();
    hs.calculate();

    mint ans = 0;
    rep(i, n) {
        if (s[i] != 'B' && s[i] != '?') continue;
        ll a = as.getSectionSum(0, i - 1);
        ll lh = hs.getSectionSum(0, i - 1);
        ll c = cs.getSectionSum(i + 1, n - 1);
        ll rh = hs.getSectionSum(i + 1, n - 1);

        mint l1 = mint(a) * mint(3).pow(lh);
        mint l2 = lh == 0 ? 0 : mint(lh) * mint(3).pow(lh - 1);

        mint r1 = mint(c) * mint(3).pow(rh);
        mint r2 = rh == 0 ? 0 : mint(rh) * mint(3).pow(rh - 1);

        mint l = l1 + l2;
        mint r = r1 + r2;

        mint now = l * r;
        ans += now;
    }

    cout << ans << endl;
}