Largest Subsequence
Given is a string $$$s$$$ of length $$$n$$$. In one operation you can select the lexicographically largest$$$^\dagger$$$ subsequence of string $$$s$$$ and cyclic shift it to the right$$$^\ddagger$$$. Your task is to calculate the minimum number of operations it would take for $$$s$$$ to become sorted, or report that it never reaches a sorted state.$$$^\dagger$$$A string $$$a$$$ is lexicographically smaller than a string $$$b$$$ if and only if one of the following holds: $$$a$$$ is a prefix of $$$b$$$, but $$$a \ne b$$$; In the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ has a letter that appears earlier in the alphabet than the corresponding letter in $$$b$$$.$$$^\ddagger$$$By cyclic shifting the string $$$t_1t_2\ldots t_m$$$ to the right, we get the string $$$t_mt_1\ldots t_{m-1}$$$.
In the first test case, the string $$$s$$$ is already sorted, so we need no operations.In the second test case, doing one operation, we will select cb and cyclic shift it. The string $$$s$$$ is now abc which is sorted.In the third test case, $$$s$$$ cannot be sorted.In the fourth test case we will perform the following operations: The lexicographically largest subsequence is zca. Then $$$s$$$ becomes abzc. The lexicographically largest subsequence is zc. Then $$$s$$$ becomes abcz. The string becomes sorted.Thus, we need $$$2$$$ operations.
Each test consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. The description of the test cases follows.The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of the string $$$s$$$.The second line of each test case contains a single string $$$s$$$ of length $$$n$$$, consisting of lowercase English letters.It is guaranteed that sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.
For each test case, output a single integer — the minimum number of operations required to make $$$s$$$ sorted, or $$$-1$$$ if it's impossible.
for s in [*open(0)][2::2]:
    (*a,) = s[:-1]
    b = []
    c = m = ''
    i = len(a)
    while i:
        i -= 1
        if a[i] >= m:
            m = a[i]
            b += (i,)
            c += a[i]
    for x in c:
        a[b.pop()] = x
    print((len(c.strip(c[-1])), -1)[a > sorted(a)])
for s in [*open(0)][2::2]:
    (*a,) = s[:-1]
    b = []
    c = m = ''
    i = len(a)
    while i:
        i -= 1
        if a[i] >= m:
            m = a[i]
            b += (i,)
            c += a[i]
    for x in c:
        a[b.pop()] = x
    print((len(c.strip(c[-1])), -1)[a > sorted(a)])
for s in [*open(0)][2::2]:
    (*a,) = s[:-1]
    b = []
    c = m = ''
    i = len(a)
    while i:
        i -= 1
        if a[i] >= m:
            m = a[i]
            b += (i,)
            c += a[i]
    for x in c:
        a[b.pop()] = x
    print((len(c.strip(c[-1])), -1)[a > sorted(a)])