Love Story
Timur loves codeforces. That's why he has a string $$$s$$$ having length $$$10$$$ made containing only lowercase Latin letters. Timur wants to know how many indices string $$$s$$$ differs from the string "codeforces".For example string $$$s =$$$ "coolforsez" differs from "codeforces" in $$$4$$$ indices, shown in bold.Help Timur by finding the number of indices where string $$$s$$$ differs from "codeforces".Note that you can't reorder the characters in the string $$$s$$$.
The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases.Each test case is one line and contains the string $$$s$$$, consisting of exactly $$$10$$$ lowercase Latin characters.
For each test case, output a single integer — the number of indices where string $$$s$$$ differs.
def count_dffer(ex_str):
    codeforces = 'codeforces'
    str_list = []
    count = 0
    for l in codeforces:
        str_list.append(l)
    for (i, s) in enumerate(ex_str):
        if codeforces[i] != ex_str[i]:
            count += 1
    print(count)
count = int(input())
for _ in range(count):
    ex_str = input()
    count_dffer(ex_str)
def main():
    t = int(input())
    for _ in range(t):
        word = str(input())
        c = 'codeforces'
        count = 0
        for i in range(0, len(word)):
            if c[i] != word[i]:
                count += 1
        print(count)
if __name__ == '__main__':
    main()
count = 0
for _ in range(int(input())):
    s = list(input())
    if s[0] != 'c':
        count += 1
    if s[1] != 'o':
        count += 1
    if s[2] != 'd':
        count += 1
    if s[3] != 'e':
        count += 1
    if s[4] != 'f':
        count += 1
    if s[5] != 'o':
        count += 1
    if s[6] != 'r':
        count += 1
    if s[7] != 'c':
        count += 1
    if s[8] != 'e':
        count += 1
    if s[9] != 's':
        count += 1
    print(count)
    count = 0