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;

using ll = long long;
using ld = long double;

using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vb = vector<bool>;
using vs = vector<string>;

typedef pair<ll, ll> P;

#define bit(n) (1LL << (n))

//#define int long long

#define all(v) v.begin(), v.end()
#define sortAl(v) sort(all(v))
#define sortAlr(v)          \
  sort(v.begin(), v.end()); \
  reverse(v.begin(), v.end())

#define rep(i, n) for (ll i = 0; i < n; i++)
#define REP(i, n) for (ll i = 1; i < n; i++)

#define FOR(i, a, b) for (ll i = (a); i < (b); i++)
#define FORm(i, m) for (auto i = m.begin(); i != m.end(); i++)

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

#define mod (ll)(1e9 + 7)
#define INF LLONG_MAX

signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);

  string s;
  cin >> s;

  ll n = s.size();

  vvll dp(4, vll(n + 1));

  if (s[0] == 'A') {
    dp[0][0] = 1;
    dp[1][0] = 1;
  }
  if (s[0] == '?') {
    dp[0][0] = 3;
    dp[1][0] = 1;
  }

  REP(i, n) {
    // 1こ前のdpを引き継ぎ
    rep(j, 4) {
      dp[j][i] = dp[j][i - 1];
      // ? なら3倍にして引き継ぐ
      if (s[i] == '?') {
        dp[j][i] *= 3;
        dp[j][i] %= mod;
      }
    }

    rep(j, 3) {
      if (s[i] == 'A' + j || s[i] == '?') {
        dp[j + 1][i] += dp[j][i - 1];
      }
    }

    rep(j, 4) {
      dp[j][i] %= mod;
    }
  }

  cout << dp[3][n - 1] << endl;

  return 0;
}