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 JavaScript (node.js v5.12) (judge verdict: Accepted)

function Main(s){
  var a = [0];
  var c = [0];
  var q = [0];
  var l = s.length;
  for(var i = 1; i <= l; i++){
    if(s[i-1] === "?"){a[i] = a[i-1]; c[i] = c[i-1]; q[i] = q[i-1] + 1;}
    else if(s[i-1] === "A"){a[i] = a[i-1] + 1; c[i] = c[i-1]; q[i] = q[i-1];}
    else if(s[i-1] === "C"){a[i] = a[i-1]; c[i] = c[i-1] + 1; q[i] = q[i-1];}
    else {a[i] = a[i-1]; c[i] = c[i-1]; q[i] = q[i-1];}
  }
  var ans = 0;
  for(var i = 2; i < l; i++){
    if(s[i-1] !== "B" && s[i-1] !== "?")continue;
    ans += abmod(a[i-1], c[l]-c[i], powmod(3,q[i-1]), powmod(3,q[l]-q[i]));
    ans += abmod(q[i-1], c[l]-c[i], powmod(3,q[i-1]-1), powmod(3,q[l]-q[i]));
    ans += abmod(a[i-1], q[l]-q[i], powmod(3,q[i-1]), powmod(3,q[l]-q[i]-1));
    ans += abmod(q[i-1], q[l]-q[i], powmod(3,q[i-1]-1), powmod(3,q[l]-q[i]-1));
    ans %= mod
  }
  console.log(ans);
}
var mod = 1e9 + 7;
function abmod(){var a=arguments,r=a[0],i=1;for(;i<a.length;)r=((r>>16)*a[i]%mod*65536+(r&65535)*a[i++])%mod;return r;}
function powmod(a,n){var r=1;while(n>0){if(n&1)r=abmod(a,r);a=abmod(a,a);n>>=1;}return r;}

Main(require("fs").readFileSync("/dev/stdin","utf8").trim());