Given a string `s` of uppercase letters, your task is to determine how many strings `t` (also uppercase) with length equal to that of `s` satisfy the followng conditions:

* `t` is lexicographical larger than `s`, and
* when you write both `s` and `t` in reverse order, `t` is still lexicographical larger than `s`.


```Haskell
For example:
solve('XYZ') = 5. They are: YYZ, ZYZ, XZZ, YZZ, ZZZ
```
String lengths are less than `5000`. Return you answer `modulo 10^9+7 (= 1000000007)`.

More examples in test cases. Good luck!

Difficulty: introductory

Starter code:

def solve(s):
	

Solutions:
def solve(s):
    r, l = 0, 0
    for c in s:
        m = ord('Z') - ord(c)
        r, l = r + m + l * m, m + l * 26
    return r % 1000000007

def solve(s, z=ord('Z')):
    a = r = 0
    for i in (z - ord(c) for c in s):
        r += (a + 1) * i
        a = a * 26 + i
        r %= 1000000007
        a %= 1000000007
    return r

def solve(s):
    M = 10 ** 9 + 7;
    z = ord('Z')
    base = z - ord('A') + 1
        
    left, total = 0, 0 
    # left-upside and total upside for the already traversed part of the list
    for c_num in [(z - ord(c)) for c in s]:
        l, t = left, total
        left = (l * base + c_num) % M
        total = ((l + 1) * c_num + t) % M
    return total

def solve(s):
    M = 10 ** 9 + 7;
    base = ord('Z') - ord('A') + 1
    num_cs = map(lambda c: 90 - ord(c), [c for c in s])
    acc = (0, 0) # ( left-upside, total upside ) for the already traversed part of the list
    for num_c in num_cs:
        (ups_left, ups_tot) = acc
        ups_left_ = (ups_left * base + num_c) % M
        ups_tot_ = ((ups_left + 1) * num_c + ups_tot) % M
        acc = (ups_left_, ups_tot_)
    return acc[1]

M=10**9+7
from functools import reduce

def solve(s):
    return reduce(lambda p,n: ((p[0]*26+(90-n))%M,((p[0]+1)*(90-n)+p[1])%M),map(ord,s),(0,0))[1]

def solve(s):
    a, pa = 0, 0
    for c in s:
        diff = ord('Z')-ord(c)
        a = a + diff + pa*diff
        pa = diff + pa*26
        a, pa = a%1000000007, pa%1000000007
    return a


def solve(s):
    sum_value, tmp = 0, 0
    for value in map(lambda x: 90 - ord(x), s):
        sum_value += value*(1+tmp)
        tmp = (26*tmp + value) % 1000000007
    return sum_value % 1000000007

def solve(s):
    greater_letters = list(map(lambda x: 90 - ord(x), s))
    sum_value, tmp = 0, 0
    for value in greater_letters:
        sum_value += value*(1+tmp)
        tmp = (26*tmp + value) % 1000000007
    return sum_value % 1000000007

from functools import lru_cache
from itertools import combinations


@lru_cache(5000)
def get_26_power(i):
    return 26**i


def solve(s):
    greater_letters = list(map(lambda x: 90 - ord(x), s))
    tmp = 1 + sum(greater*get_26_power(i) for i, greater in enumerate(greater_letters[1:]))
    sum_value = 0
    for value in greater_letters:
        sum_value += tmp*value
        tmp = (tmp+25)//26
    return sum_value % 1000000007

def solve(s):
    val = [90 - ord(c) for c in reversed(s)]
    tmp, res = [0], val[0]
    for v1,v2 in zip(val, val[1:]):
        tmp.append(tmp[-1]*26 + v1)
        res += v2 * (tmp[-1]+1)
    return res % (10**9+7)

MOD = pow(10, 9) + 7

def solve(s):
    n = len(s)
    a = [25 - (ord(x) - 65) for x in s]
    r, s = 0, 0
    for i in range(n):
        r = (r + a[i] + s * a[i]) % MOD
        s = (a[i] + s * 26) % MOD
    return r

def solve(s):
    n = len(s); dp, mod, sum = [0]*n, 1000000007, 0
    for i in range(n-2,-1,-1):
        dp[i] = (((dp[i + 1] * 26) % mod) + (ord('Z') - ord(s[i + 1])) % mod) % mod
    for i in range(0,n):
        sum += ((dp[i] + 1) * (ord('Z') - ord(s[i]))) % mod
        sum %= mod
    return sum