Let (f(s)) be the compressed version of a string (s), formed by replacing every maximal contiguous block of identical characters with a single copy of that character. For example, (f()" aabbcc "() \ = \)" abc ". Let (|s|) denote the length of a string (s). Following this, (|f(s)|) denotes the length of the compressed string. For example: (|f()" aabbcc "()| =) (|)" abc "(|) (= 3) If the string is empty, its length is (0). Yousef has given you a string (s) consisting of (n) lowercase Latin letters. You must delete exactly one character (s_i) ((2 \le i \le n - 1)) to form a new string (s'), and then find the minimum possible value of (|f(s')|). Note that you cannot delete (s_1) or (s_n). The first line contains an integer (t) ((1 \le t \le 10^4)) — the number of test cases. The first line of each test case contains an integer (n) ((3 \le n \le 2 \cdot 10^5)) — the length of the string. The second line of each test case contains a string (s) ((|s| = n)), consisting of lowercase Latin letters. It is guaranteed that the sum of (n) over all test cases does not exceed (2 \cdot 10^5). For each test case, output a single integer — the minimum possible length of the resulting compressed string after deleting one character. In the first test case, we can only delete the character (s_2 =) ' b ', producing a string (s' =) " ab ", with (|f(s')| = 2). Therefore, (2) is the minimum length achievable. In the fourth test case, we can delete the character (s_2 =) ' b '. The resulting string is (s' =) " aaa " with (f(s') =) " a ", so (|f(s')| = 1). In the sixth test case, deleting any valid character results in (f(s') =) " e " and (|f(s')| = 1). In the eighth test case, we can delete the character (s_4 =) ' c '. The resulting string is (s' =) " abaaba " with (f(s') =) " ababa " and (|f(s')| = 5).