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 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<string.h>
long long int modpow(long long int a, long long int n, long long int p)
{
	if (n < 0)return 0;
	long long int res = 1;
	while (n > 0)
	{
		if (n % 2 == 1)
			res = res * a % p;
		a = a * a % p;
		n /= 2;
	}
	return res;
}
int main()
{
	char s[100005];
	scanf("%s", s);
	long long int i, n = strlen(s);
	long long int ca[100005], cha[100005];
	ca[0] = cha[0] = 0;
	for (i = 0; i < n; i++)
	{
		if (s[i] == 'A')
		{
			ca[i + 1] = ca[i] + 1;
			cha[i + 1] = cha[i];
		}
		else if (s[i] == '?')
		{
			ca[i + 1] = ca[i];
			cha[i + 1] = cha[i] + 1;
		}
		else
		{
			ca[i + 1] = ca[i];
			cha[i + 1] = cha[i];
		}
	}
	long long int cc[100005], chc[100005];
	cc[n] = chc[n] = 0;
	for (i = n - 1; i >= 0; i--)
	{
		if (s[i] == 'C')
		{
			cc[i] = cc[i + 1] + 1;
			chc[i] = chc[i + 1];
		}
		else if (s[i] == '?')
		{
			cc[i] = cc[i + 1];
			chc[i] = chc[i + 1] + 1;
		}
		else
		{
			cc[i] = cc[i + 1];
			chc[i] = chc[i + 1];
		}
	}
	long long int p = 1000000007;
	long long int ans = 0;
	for (i = 0; i < n; i++)
		if (s[i] == 'B' || s[i] == '?')
			ans = (ans + ((ca[i] * modpow(3, cha[i], p) % p + cha[i] * modpow(3, cha[i] - 1, p) % p) % p) * ((cc[i + 1] * modpow(3, chc[i + 1], p) % p + chc[i + 1] * modpow(3, chc[i + 1] - 1, p) % p) % p)) % p;
	printf("%lld\n", ans);
	return 0;
}