def has_three_o(s):
    for i in range(len(s) - 2):
        if s[i] == 'o' and s[i+1] == 'o' and s[i+2] == 'o':
            return True
    return False

T = int(input())
for _ in range(T):
    N, S = input().split()
    S = S.strip()
    if has_three_o(S):
        print("O")
        continue
    c = S.count('-')
    m = (c + 1) // 2  # O's remaining moves
    possible = False
    for i in range(len(S) - 2):
        a = 0
        b = 0
        for j in range(3):
            if S[i + j] == 'o':
                a += 1
            elif S[i + j] == '-':
                b += 1
        needed = max(0, 3 - a)
        ceil_b = (b + 1) // 2  # ceil(b/2)
        if min(m, ceil_b) >= needed:
            possible = True
            break
    print("O" if possible else "X")