# 106正式試題

## pA. 計算含有 s 或 S 字母的字數

for _ in range(int(input())):
    a = input().split()
    ans = 0
    for i in a:
        if "S" in i or "s" in i:
            ans += 1
    print(ans)

## pB. 給一個羅馬數字符號，轉為整數數字

key = {"I":1, "V":5, "X":10, "L":50, "C":100, "D":500, "M":1000}
for _ in range(int(input())):
    n = input()
    pre = 0
    ans = 0
    for i in n:
        value = key[i]
        if value > pre:
            ans += value - 2*pre
        else:
            ans += value
        pre = value
    print(ans)

## pC. 信用卡卡號

for _ in range(int(input())):
    n = input()
    ans = 0
    for i in range(len(n)):
        i = int(i)
        if i%2:
            ans += int(n[i])
        else:
            if int(n[i])*2 >= 10:
                ans += int(str(int(n[i])*2)[1])+1
            else:
                ans += int(n[i])*2
    print("F" if ans%10 else "T")

## pD. 幾 A 幾 B

def dfs(n, d):
    if len(n) == len(d): now.add(d)
    for i in n:
        if i not in d:
            dfs(n, d+i)
for _ in range(int(input())):
    n, i, j = map(int, input().split(","))
    now = set()
    dfs(str(n), "")
    now = sorted(now)
    i = now[i-1]
    j = now[j-1]
    A, B = 0, 0
    for v in i:
        if v in j:
            if i.index(v) == j.index(v):
                A += 1
            else:
                B += 1
    print(f"{A}A{B}B")

## pE. 網段廣播位址

for _ in range(int(input())):
    a, b = input().split("/")
    a = [int(i) for i in a.split(".")]
    b = [int(i) for i in b.split(".")]
    for i in range(len(a)):
        a[i] = a[i] | ~b[i]+256
    print(".".join(map(str, a)))

## pF. 大數排序問題

for _ in range(int(input())):
    num = [int(i) for i in input().split(", ")]
    n = sorted(num.copy())
    for i in range(1, len(num)+1):
        num[num.index(n[i-1])] = i
    print(", ".join(map(str, num)))

## pG. 樹
ChatGPT

from collections import defaultdict

# 建立鄰接表
def build_graph(edges):
    graph = defaultdict(list)
    for u, v in edges:
        graph[u].append(v)
        graph[v].append(u)
    return graph

# 檢查無向圖是否有迴圈，並輸出結果
def check_graph(edges):
    graph = build_graph(edges)
    visited = set()
    cycle_nodes = []
    is_connected = True  # 檢查是否為單連通圖
    start_node = next(iter(graph))  # 任意選擇起始節點

    def dfs(node, parent, path):
        visited.add(node)
        path.append(node)
        
        for neighbor in graph[node]:
            if neighbor not in visited:
                if dfs(neighbor, node, path):
                    return True
            elif neighbor != parent:
                # 形成了迴圈，記錄迴圈節點
                cycle_start_index = path.index(neighbor)
                cycle_nodes.extend(path[cycle_start_index:])
                return True
        path.pop()
        return False

    # 若找到迴圈則返回迴圈節點
    if dfs(start_node, -1, []):
        return sorted(cycle_nodes)

    # 若無迴圈，確認是否為單連通圖
    if len(visited) != len(graph):
        return "F"
    return "T"

# 測試範例
for _ in range(int(input())):
    edges = [eval(i) for i in input().split()]
    result = check_graph(edges)
    if type(result) == list:
        print(", ".join(map(str, result)))
    else:
        print(result)

## pH. 後序運算式

for _ in range(int(input())):
  data = list(input().split())[::-1]
  stack = []
  while data != []:
    num_or_op = data.pop()
    try:#int
      eval(num_or_op)
      stack.append(num_or_op)
    except:#op
      b = stack.pop()
      a = stack.pop()
      stack.append(str(eval(a+num_or_op+b)))
  print(format(float(stack[0]),".0f"))