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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#pragma GCC optimize("Ofast", "unroll-loops")
using namespace std;
#define ll long long
#define int ll
#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define REP(i,n) FOR(i,0,n)
#define REP1(i,n) FOR(i,1,n+1)
#define RREP(i,n) for (int i=(n)-1;i>=0;i--)
#define pii pair<int, int>
#define f first
#define s second
#define pb push_back
#define ALL(x) x.begin(), x.end()
#define SZ(x) (int)(x.size())
 
#ifdef TOAD
#define IOS()
#define bug(x) cerr<<"Line "<<__LINE__<<": "<<#x<<" is "<<x<<endl
#else 
#define IOS() ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#define bug(x)
#endif 
const ll inf=(1ll<<60);
const double PI=acos(-1);
const int iinf=2147483647;
const ll mod=1e9+7;
const ll maxn=1e5+5;
ll pw(ll x, ll p){
	ll ret=1;
	while(p>0){
		if (p&1){
			ret*=x;
			ret%=mod;
		}
		p>>=1;
		x*=x;
		x%=mod;
	}
	return ret;
}
int op(int a, int qq){
	if (qq==0){
		return a;
	}
	else if (qq==1){
		return (a*3+1)%mod;
	}
	int x=a*pw(3,qq)%mod;
	int y=pw(3,qq-1)*qq%mod;
	return (x+y)%mod;
}
signed main(){
	IOS();
	string str; cin>>str;
    int n=SZ(str);
	int sum=0;
	int as=0, qs=0;
	vector<pii> vc(n);
	RREP(i,n){
		if (i==n-1) {
			vc[i]={0,0};
		}
		else{
			vc[i]={vc[i+1].f+(str[i+1]=='C'), vc[i+1].s+(str[i+1]=='?')};
		}
	}
	FOR(i,0,n){
		//cout<<vc[i].f<<' '<<vc[i].s<<endl;
 
		int va=op(as, qs);
		int vb=op(vc[i].f, vc[i].s);
		if (str[i]=='?'||str[i]=='B') sum+=va*vb%mod;
		sum%=mod;
		if (str[i]=='A'){
			as++;
		}
		else if (str[i]=='?') qs++;
		
	}
	cout<<sum<<endl;
}