You are given a string s consisting of lowercase Latin letters "a", "b" and "c" and question marks "?".

Let the number of question marks in the string s be k. Let's replace each question mark with one of the letters "a", "b" and "c". Here we can obtain all 3^{k} possible strings consisting only of letters "a", "b" and "c". For example, if s = "ac?b?c" then we can obtain the following strings: ["acabac", "acabbc", "acabcc", "acbbac", "acbbbc", "acbbcc", "accbac", "accbbc", "accbcc"].

Your task is to count the total number of subsequences "abc" in all resulting strings. Since the answer can be very large, print it modulo 10^{9} + 7.

A subsequence of the string t is such a sequence that can be derived from the string t after removing some (possibly, zero) number of letters without changing the order of remaining letters. For example, the string "baacbc" contains two subsequences "abc" — a subsequence consisting of letters at positions (2, 5, 6) and a subsequence consisting of letters at positions (3, 5, 6).

Input

The first line of the input contains one integer n (3 ≤ n ≤ 200 000) — the length of s.

The second line of the input contains the string s of length n consisting of lowercase Latin letters "a", "b" and "c" and question marks"?".

Output

Print the total number of subsequences "abc" in all strings you can obtain if you replace all question marks with letters "a", "b" and "c", modulo 10^{9} + 7.

Examples

Input


6
ac?b?c


Output


24


Input


7
???????


Output


2835


Input


9
cccbbbaaa


Output


0


Input


5
a???c


Output


46

Note

In the first example, we can obtain 9 strings:

  * "acabac" — there are 2 subsequences "abc", 
  * "acabbc" — there are 4 subsequences "abc", 
  * "acabcc" — there are 4 subsequences "abc", 
  * "acbbac" — there are 2 subsequences "abc", 
  * "acbbbc" — there are 3 subsequences "abc", 
  * "acbbcc" — there are 4 subsequences "abc", 
  * "accbac" — there is 1 subsequence "abc", 
  * "accbbc" — there are 2 subsequences "abc", 
  * "accbcc" — there are 2 subsequences "abc". 



So, there are 2 + 4 + 4 + 2 + 3 + 4 + 1 + 2 + 2 = 24 subsequences "abc" in total.
"""
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                         ppppp  ppppppppppppppppppp
                      ppppppp    ppppppppppppppppppppp
                      pppppppp  pppppppppppppppppppppp
                      pppppppppppppppppppppppppppppppp
                              pppppppppppppppppppppppp
       ppppppppppppppppppppppppppppppppppppppppppppppp  pppppppppppppppppppp
      pppppppppppppppppppppppppppppppppppppppppppppppp  ppppppppppppppppppppp
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   pppppppppppppppppppppppppppppppppppppppppppppp     pppppppppppppppppppppppppp
  ppppppppppppppppppppppppppppppppppppppppppppp      pppppppppppppppppppppppppppp
  pppppppppppppppppppppppppppppppp               pppppppppppppppppppppppppppppppp
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  ppppppppppppppppppppppppppp    pppppppppppppppppppppppppppppppppppppppppppppppp
    pppppppppppppppppppppppp  pppppppppppppppppppppppppppppppppppppppppppppppppp
     ppppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppppppp
      pppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppppp
       ppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppp
                              pppppppppppppppppppppppp
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                              pppppppppppppppppppppp  pppppppp
                              ppppppppppppppppppppp    ppppppp
                                 ppppppppppppppppppp  ppppp
                                    pppppppppppppppppppp
"""


import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush, nsmallest
from math import ceil, floor, gcd, fabs, factorial, fmod, sqrt, inf
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
from decimal import Decimal
# sys.setrecursionlimit(pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(var): sys.stdout.write(str(var)+"\n")
def outa(*var, end="\n"): sys.stdout.write(' '.join(map(str, var)) + end)
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]


n = int(data())
s = data()
a, b, c = 0, 0, 0
temp = 1
for i in s:
    if i == 'a':
        a = (a + temp) % mod
        continue
    if i == 'b':
        b = (b + a) % mod
        continue
    if i == 'c':
        c = (c + b) % mod
        continue
    c = (c * 3 + b) % mod
    b = (b * 3 + a) % mod
    a = (a * 3 + temp) % mod
    temp = (temp * 3) % mod
out(c)
