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: Accepted)

#include <bits/stdc++.h>
#define sp ' '
#define nyan "(=^・ω・^=)"
#define mkp make_pair
#define intmax 2147483647
#define llmax 9223372036854775807
#define lP pair<ll,ll>
#define iP pair<int,int>
typedef long long ll;
using namespace std;
const int mod = 1000000007;

ll cntA[100001], cntB[100001], cntC[100001], cntQ[100001];
string s;
ll a[4], res;

ll pw(ll x, ll y) {
	ll a = 1;
	while (y) {
		if (y & 1) {
			a *= x;
			a %= mod;
		}
		x *= x;
		x %= mod;
		y /= 2;
	}
	return a;
}

int main() {
	cin >> s;
	for (int i = 0; i != s.size(); ++i) {
		cntA[i + 1] = cntA[i];
		cntB[i + 1] = cntB[i];
		cntC[i + 1] = cntC[i];
		cntQ[i + 1] = cntQ[i];
		if (s[i] == 'A')
			++cntA[i + 1];
		if (s[i] == 'B')
			++cntB[i + 1];
		if (s[i] == 'C')
			++cntC[i + 1];
		if (s[i] == '?')
			++cntQ[i + 1];
	}
	for (int i = 0; i != s.size(); ++i) {
		if (s[i] == 'B') {
			a[3] += cntA[i] * (cntC[s.size()] - cntC[i + 1]);
			a[3] %= mod;
			a[2] += cntA[i] * (cntQ[s.size()] - cntQ[i + 1]);
			a[2] %= mod;
			a[2] += cntQ[i] * (cntC[s.size()] - cntC[i + 1]);
			a[2] %= mod;
			a[1] += cntQ[i] * (cntQ[s.size()] - cntQ[i + 1]);
			a[1] %= mod;
		}
		if (s[i] == '?') {
			a[2] += cntA[i] * (cntC[s.size()] - cntC[i + 1]);
			a[2] %= mod;
			a[1] += cntA[i] * (cntQ[s.size()] - cntQ[i + 1]);
			a[1] %= mod;
			a[1] += cntQ[i] * (cntC[s.size()] - cntC[i + 1]);
			a[1] %= mod;
			a[0] += cntQ[i] * (cntQ[s.size()] - cntQ[i + 1]);
			a[0] %= mod;
		}
	}
	res += a[3] * pw(3, cntQ[s.size()]);
	res %= mod;
	res += a[2] * pw(3, cntQ[s.size()] - 1);
	res %= mod;
	res += a[1] * pw(3, cntQ[s.size()] - 2);
	res %= mod;
	res += a[0] * pw(3, cntQ[s.size()] - 3);
	res %= mod;
	cout << res << endl;
	return 0;
}