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 <iostream>
#include <string>
#include <cstring> 

using namespace std;

#ifdef DEBUG_MODE
#define DBG(n) n;
#else
#define DBG(n) ;
#endif
#define REP(i,n) for(ll (i) = (0);(i) < (n);++i)
#define PB push_back
#define MP make_pair
#define FI first
#define SE second
#define SHOW1d(v,n) {for(int W = 0;W < (n);W++)cerr << v[W] << ' ';cerr << endl << endl;}
#define SHOW2d(v,i,j) {for(int aaa = 0;aaa < i;aaa++){for(int bbb = 0;bbb < j;bbb++)cerr << v[aaa][bbb] << ' ';cerr << endl;}cerr << endl;}
#define ALL(v) v.begin(),v.end()
#define Decimal fixed<<setprecision(20)
#define INF 1000000000
#define MOD 1000000007

typedef long long ll;

long long mod_div(long long a, long long b) {
	if (a % b == 0)return a / b;
	long long tmp = MOD - 2, c = b, ret = 1;
	while (tmp > 0) {
		if (tmp & 1) {
			ret *= c; ret %= MOD;
		}
		c *= c; c %= MOD; tmp >>= 1;
	}
	return a*ret%MOD;
}

char str[100011];
ll dp[4][100011];

int main() {

	scanf("%s", str);

	ll tmp = 1;
	ll cou = 0;
	REP(i, strlen(str)) {
		if (str[i] == '?')cou++;
	}
	REP(i, cou) {
		tmp *= 3;
		tmp %= MOD;
	}

	dp[0][0] = tmp;
	REP(i, strlen(str)) {
		REP(j, 4) {
			if (dp[j][i]) {
				if (j < 3) {
					if (str[i] == 'A' + j) {
						dp[j + 1][i + 1] += dp[j][i];
						dp[j + 1][i + 1] %= MOD;
					}
					else if (str[i] == '?') {
						dp[j + 1][i + 1] += mod_div(dp[j][i], 3);
						dp[j + 1][i + 1] %= MOD;
					}
				}
				dp[j][i + 1] += dp[j][i];
				dp[j][i + 1] %= MOD;
			}
		}
	}

	cout << dp[3][strlen(str)] << endl;

	return 0;
}