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.

Examples

Input

A??C


Output

8


Input

ABCBC


Output

3


Input

????C?????B??????A???????


Output

979596887
#include <stdio.h>

#define DIV 1000000007

char S[100008];
int Snum=0;

long long dp[100008][4];

int main()
{
    scanf("%s", S);
    for (Snum = 0; S[Snum]!='\0'; Snum++);
    
    dp[Snum][3]=1;
    dp[Snum][2]=0;
    dp[Snum][1]=0;
    dp[Snum][0]=0;
    
    for(int i=Snum-1;i>=0;i--){
        bool isq=S[i]=='?'?true:false;
        dp[i][3]=(isq?3:1)*dp[i+1][3]%DIV;
        //printf("%d",dp[i][3]);
        for(int j=2;j>=0;j--){
            dp[i][j]=((isq?3:1)*dp[i+1][j]
            +((isq||S[i]-'A'==j)?1:0)*dp[i+1][j+1])%DIV;
            //printf(",%d",dp[i][j]);
        }
        //printf("\n");
    }
    printf("%lld\n",dp[0][0]);

    return 0;
}