The notorious hacker group "Sed" managed to obtain a string $S$ from their secret sources. The string contains only lowercase English letters along with the character '?'.
A substring of $S$ is a contiguous subsequence of that string. For example, the string "chef" is a substring of the string "codechef", but the string "codeh" is not a substring of "codechef".
A substring of $S$ is good if it is possible to choose a lowercase English letter $C$ such that the following process succeeds:
- Create a string $R$, which is a copy of the substring, but with each '?' replaced by the letter $c$. Note that all occurrences of '?' must be replaced by the same letter.
- For each lowercase English letter:
- Compute the number of times it occurs in $S$ and the number of times it occurs in $R$; let's denote them by $A$ and $B$ respectively. The '?' characters in the original string $S$ are ignored when computing $A$.
- Check the parity of $A$ and $B$. If they do not have the same parity, i.e. one of them is even while the other is odd, then this process fails.
- If the parities of the number of occurrences in $S$ and $R$ are the same for each letter, the process succeeds.
For different substrings, we may choose different values of $C$.
Help Sed find the number of good substrings in the string $S$.

-----Input-----
- The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
- The first and only line of each test case contains a single string $S$.

-----Output-----
For each test case, print a single line containing one integer ― the number of good substrings.

-----Constraints-----
- $1 \le T \le 10^5$
- $1 \le S \le 10^5$
- $S$ contains only lowercase English letters ('a' through 'z') and '?'
- the sum of $|S|$ over all test cases does not exceed $10^5$

-----Example Input-----
5
aa?
a???
????
asfhaslskfak
af??avvnfed?fav?faf????

-----Example Output-----
2
6
4
2
27

-----Explanation-----
Example case 1: All letters occur an even number of times in $S$. The six substrings of $S$ are "a", "a", "?", "aa", "a?" and "aa?" (note that "a" is counted twice). Which of them are good?
- "a" is not good as 'a' occurs an odd number of times in this substring and there are no '?' to replace.
- "?" is also not good as replacing '?' by any letter causes this letter to occur in $R$ an odd number of times.
- "aa" is a good substring because each letter occurs in it an even number of times and there are no '?' to replace.
- "a?" is also a good substring. We can replace '?' with 'a'. Then, $R$ is "aa" and each letter occurs in this string an even number of times. Note that replacing '?' e.g. with 'b' would not work ($R$ would be "ab", where both 'a' and 'b' occur an odd number of times), but we may choose the replacement letter $C$ appropriately.
- "aa?" is not a good substring. For any replacement letter $C$, we find that $C$ occurs an odd number of times in $R$.
Example case 2: We especially note that "a???" is not a good substring. Since all '?' have to be replaced by the same character, we can only get strings of the form "aaaa", "abbb", "accc", etc., but none of them match the criteria for a good substring.
Example case 3: Any substring with even length is good.

Difficulty: VERY_HARD
Tags: Bit manipulation

Constructive algorithms

String algorithms

Mathematics

Data structures

Ad-hoc
Skill types: Bit manipulation

Data structures

Starter code:

Solutions:
from sys import stdin, stdout
input = stdin.readline

def val(i):
	if i == '?':
		return 26
	return ord(i) - 97
t = int(input())
for _ in range(t):
	s = input()[:-1]
	ans = 0
	mask = 0
	for i in s:
		if i == '?':
			continue
		mask ^= 1 << val(i)
	hash = {0: 1}
	pre = 0
	for i in s:
		pre ^= 1 << val(i)
		ans += hash.get(pre ^ mask, 0)
		x = mask ^ 1 << 26
		for j in range(26):
			ans += hash.get(pre ^ 1 << j ^ x, 0)
		hash[pre] = hash.get(pre, 0) + 1
	stdout.write(str(ans) + '\n')


def to_bit(c):
	if c == '?':
		bit = 26
	else:
		bit = ord(c) - ord('a')
	return bit
T = int(input())
for _ in range(T):
	s = input()
	mask = 0
	for c in s:
		if c != '?':
			mask ^= 1 << to_bit(c)
	good_prefixes = []
	good_prefixes.append(mask & (1 << to_bit('?')) - 1)
	for i in range(26):
		good_prefixes.append(mask ^ 1 << i | 1 << to_bit('?'))
	prefixes = {0: 1}
	res = 0
	current = 0
	for c in s:
		current ^= 1 << to_bit(c)
		for prefix in good_prefixes:
			res += prefixes.get(prefix ^ current, 0)
		prefixes[current] = prefixes.get(current, 0) + 1
	print(res)


def convertToParitys(s):
	keys = '?abcdefghijklmnopqrstuvwxyz'
	paritys = {c: 0 for c in keys}
	for c in s:
		paritys[c] += 1
	for (c, v) in paritys.items():
		paritys[c] = v % 2
	out = 0
	bitValue = 1
	for c in keys:
		if paritys[c]:
			out += bitValue
		bitValue *= 2
	return out

def getSolutionBitMaps(s):
	out = []
	sP = convertToParitys(s)
	if sP % 2:
		sP -= 1
	out.append(sP)
	for n in range(1, 27):
		out.append(sP ^ 1 + 2 ** n)
	return out

def getLeadingSubStringBitMapCounts(s):
	out = {}
	bM = 0
	keys = '?abcdefghijklmnopqrstuvwxyz'
	paritys = {c: 0 for c in keys}
	values = {c: 2 ** i for (i, c) in enumerate(keys)}
	out[bM] = out.setdefault(bM, 0) + 1
	bMis = []
	i = 0
	bMis = [0]
	for c in s:
		i += 1
		if paritys[c]:
			paritys[c] = 0
			bM -= values[c]
		else:
			paritys[c] = 1
			bM += values[c]
		out[bM] = out.setdefault(bM, 0) + 1
		bMis.append(bM)
	return (out, bMis)

def solve(s):
	out = 0
	(bMjCounts, bMis) = getLeadingSubStringBitMapCounts(s)
	solutions = getSolutionBitMaps(s)
	for bMi in bMis:
		for bMs in solutions:
			if bMs ^ bMi in bMjCounts:
				out += bMjCounts[bMs ^ bMi]
	if 0 in solutions:
		out -= len(s)
	out //= 2
	return out
T = int(input())
for tc in range(T):
	s = input()
	print(solve(s))


def letterNo(c):
	if i == '?':
		return 26
	return ord(c) - 97
for _ in range(int(input())):
	s = input().strip()
	mask = 0
	for i in s:
		if i != '?':
			mask ^= 1 << letterNo(i)
	d = {}
	x = 0
	d[0] = 1
	ans = 0
	for i in s:
		x ^= 1 << letterNo(i)
		ans += d.get(x ^ mask, 0)
		for j in range(26):
			ans += d.get(x ^ 1 << 26 ^ 1 << j ^ mask, 0)
		d[x] = d.get(x, 0) + 1
	print(ans)


import sys

def solve(S):
	N = len(S)
	parity_S = 0
	qs_add = 1 << 26
	ord_a = ord('a')
	for (i, c) in enumerate(S):
		if c != '?':
			parity_S ^= 1 << ord(c) - ord_a
	good_Rs = [parity_S]
	for c in range(26):
		good_Rs.append(parity_S ^ 1 << c | qs_add)
	prefix_parities = dict()
	prefix_parities[0] = 1
	parity_R = 0
	soln = 0
	for (j, c) in enumerate(S):
		if c == '?':
			parity_R ^= qs_add
		else:
			parity_R ^= 1 << ord(c) - ord_a
		for good_R in good_Rs:
			soln += prefix_parities.get(good_R ^ parity_R, 0)
		if parity_R in prefix_parities:
			prefix_parities[parity_R] += 1
		else:
			prefix_parities[parity_R] = 1
	return soln

def main():
	test_cases = int(sys.stdin.readline())
	for _ in range(test_cases):
		S = sys.stdin.readline().strip()
		print(solve(S))
main()


from sys import stdin, stdout
from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log
from collections import defaultdict as dd, deque
from heapq import merge, heapify, heappop, heappush, nsmallest
from bisect import bisect_left as bl, bisect_right as br, bisect
mod = pow(10, 9) + 7
mod2 = 998244353

def inp():
	return stdin.readline().strip()

def out(var, end='\n'):
	stdout.write(str(var) + '\n')

def outa(*var, end='\n'):
	stdout.write(' '.join(map(str, var)) + end)

def lmp():
	return list(mp())

def mp():
	return map(int, inp().split())

def smp():
	return map(str, inp().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)]

def remadd(x, y):
	return 1 if x % y else 0

def ceil(a, b):
	return (a + b - 1) // b

def isprime(x):
	if x <= 1:
		return False
	if x in (2, 3):
		return True
	if x % 2 == 0:
		return False
	for i in range(3, int(sqrt(x)) + 1, 2):
		if x % i == 0:
			return False
	return True

def xor(a, b):
	c = ''
	for i in range(27):
		if a[i] == '0' and b[i] == '0' or (a[i] == '1' and b[i] == '1'):
			c += '0'
		else:
			c += '1'
	return c
for _ in range(int(inp())):
	s = inp()
	l = len(s)
	md = {}
	S = 'abcdefghijklmnopqrstuvwxyz'
	for i in S:
		md[i] = 0
	cq = 0
	mdd = {}
	pre = ['0' * 27]
	for i in range(l):
		if s[i] in md:
			md[s[i]] += 1
		b = ''
		for j in S:
			b += '1' if md[j] % 2 else '0'
		if s[i] == '?':
			cq += 1
		k = '1' if cq % 2 else '0'
		kk = b + k
		pre.append(kk)
		if kk in mdd:
			mdd[kk].append(i + 1)
		else:
			mdd[kk] = [i + 1]
	b = pre[-1]
	good = []
	good.append(b[:-1] + '0')
	b = b[:-1]
	for i in range(26):
		x = '1' if b[i] == '0' else '0'
		good.append(b[:i] + x + b[i + 1:] + '1')
	alpha = {}
	S = 'abcdefghijklmnopqrstuvwxyz?'
	for i in range(27):
		alpha[S[i]] = i
	ans = 0
	for j in good:
		prev = j
		for i in range(l + 1):
			if i != 0:
				x = alpha[s[i - 1]]
				kk = '0' if prev[x] == '1' else '1'
				prev = prev[:x] + kk + prev[x + 1:]
			if prev in mdd:
				ans += len(mdd[prev]) - br(mdd[prev], i)
	out(ans)


from sys import stdin, stdout
input = stdin.readline
t = int(input())
for _ in range(t):
	s = input()[:-1]
	n = len(s)
	mask = [0 for i in range(26)]
	for i in s:
		if i == '?':
			continue
		mask[ord(i) - 97] ^= 1
	good = [mask + [0]]
	for i in range(26):
		temp = mask.copy()
		temp[i] ^= 1
		good.append(temp + [1])
	pre_masks = {'0' * 27: 1}
	temp = [0 for i in range(27)]
	ans = 0
	for i in range(n):
		if s[i] == '?':
			temp[26] ^= 1
		else:
			temp[ord(s[i]) - 97] ^= 1
		good1 = mask + [0]
		xor = ''
		for j in range(27):
			xor += str(temp[j] ^ good1[j])
		if xor in pre_masks:
			ans += pre_masks[xor]
		good2 = mask + [1]
		xor1 = ''
		for j in range(27):
			xor1 += str(temp[j] ^ good2[j])
		for j in range(26):
			xor2 = xor1[:j] + str(int(xor1[j]) ^ 1) + xor1[j + 1:]
			if xor2 in pre_masks:
				ans += pre_masks[xor2]
		pre = ''.join([str(e) for e in temp])
		if pre in pre_masks:
			pre_masks[pre] += 1
		else:
			pre_masks[pre] = 1
	stdout.write(str(ans) + '\n')