### BUGGY CODE ###

import bisect
import cmath
import heapq
import itertools
import math
import operator
import os
import re
import string
import sys
from collections import Counter, deque, defaultdict
from copy import deepcopy
from decimal import Decimal
from fractions import gcd
from functools import lru_cache, reduce
from operator import itemgetter, mul, add, xor

import numpy as np

if os.getenv("LOCAL"):
    sys.stdin = open("_in.txt", "r")

sys.setrecursionlimit(10 ** 9)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7
# MOD = 998244353


S = sys.stdin.buffer.readline().decode().rstrip()

a = 0
ab = 0
abc = 0
q = 0
for i, s in enumerate(S):
    if s == 'C':
        abc += ab
    if s == 'B':
        ab += a
    if s == 'A':
        a += 3 ** q
    if s == '?':
        abc = abc * 3 + ab
        ab = ab * 3 + a
        a = a * 3 + 3 ** q
        q += 1
    a %= MOD
    ab %= MOD
    abc %= MOD
    # print(s, a, ab, abc)
print(abc)


### FIXED CODE ###

import functools
import os
import sys

sys.setrecursionlimit(10000)
INF = float('inf')


def debug(fn):
    if not os.getenv('LOCAL'):
        return fn

    @functools.wraps(fn)
    def wrapper(*args, **kwargs):
        ret = fn(*args, **kwargs)
        print('DEBUG: {}({}) -> '.format(
            fn.__name__,
            ', '.join(
                list(map(str, args)) +
                ['{}={}'.format(k, str(v)) for k, v in kwargs.items()]
            )
        ), end='')
        print(ret)
        return ret

    return wrapper


# @debug
# @functools.lru_cache(maxsize=None)
# def solve(idx, abc=3):
#     """
#     S[0:idx+1] の ABC 数の数
#     abc=2 なら AB 数、abc=1 なら A 数、abc=0 なら今までのパターン数
#     :param idx:
#     :param abc:
#     :return:
#     """
#     if idx < 0:
#         return 1 if abc == 0 else 0
#     c = S[idx]
#     if 1 <= abc <= 3:
#         if c == '0ABC'[abc]:
#             return (solve(idx - 1, abc=abc) + solve(idx - 1, abc=abc - 1)) % MOD
#         elif c == '?':
#             return (solve(idx - 1, abc=abc) * 3 + solve(idx - 1, abc=abc - 1)) % MOD
#         else:
#             return (solve(idx - 1, abc=abc)) % MOD
#     elif abc == 0:
#         if c == '?':
#             return (solve(idx - 1, abc=0) * 3) % MOD
#         else:
#             return (solve(idx - 1, abc=0)) % MOD


S = input()
MOD = 10 ** 9 + 7
# print(solve(len(S) - 1) % MOD)

# ↑の DP 版
# dp[i][3]: S[0:i] の ABC 数
# dp[i][2]: S[0:i] の AB 数
# dp[i][1]: S[0:i] の A 数
# dp[i][0]: S[0:i] の組み合わせ数
# dp = np.empty((len(S) + 1, 4))
# dp[0][0] = 1
# for i in range(len(S)):
#     c = S[i]
#     if c == '?':
#         dp[i + 1][0] = dp[i][0] * 3 % MOD
#     else:
#         dp[i + 1][0] = dp[i][0] % MOD
#     for abc in [1, 2, 3]:
#         if c == '0ABC'[abc]:
#             dp[i + 1][abc] = (dp[i][abc] + dp[i][abc - 1]) % MOD
#         elif c == '?':
#             dp[i + 1][abc] = (dp[i][abc] * 3 + dp[i][abc - 1]) % MOD
#         else:
#             dp[i + 1][abc] = dp[i][abc] % MOD
#     # dp[i + 1, :] %= MOD
# print(int(dp[len(S), 3]))


# ↑の配列再利用版
dp = [0] * 4
dp[0] = 1
for i in range(len(S)):
    c = S[i]
    for abc in [3, 2, 1]:
        if c == '0ABC'[abc]:
            dp[abc] = (dp[abc] + dp[abc - 1]) % MOD
        elif c == '?':
            dp[abc] = (dp[abc] * 3 + dp[abc - 1]) % MOD
    if c == '?':
        dp[0] = dp[0] * 3 % MOD
print(int(dp[3]))