Problem:
Score : 200 points

Problem Statement

The word internationalization is sometimes abbreviated to i18n.
This comes from the fact that there are 18 letters between the first i and the last n.

You are given a string s of length at least 3 consisting of lowercase English letters.
Abbreviate s in the same way.

Constraints

3 ≤ |s| ≤ 100 (|s| denotes the length of s.)

s consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

s

Output

Print the abbreviation of s.

Sample Input 1

internationalization

Sample Output 1

i18n

Sample Input 2

smiles

Sample Output 2

s4s

Sample Input 3

xyz

Sample Output 3

x1z

Solution in Rust (1.15.1) (judge verdict: Accepted)



fn main() {
    let s: String = read();
    let cs: Vec<char> = s.chars().collect();
    let n = cs.len();
    print!("{}", cs[0]);
    print!("{}", n-2);
    println!("{}", cs[n-1]);
}

// 以下関数


use std::io::*;
use std::str::FromStr;
pub fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

pub const MOD: u64 = 1_000_000_007;

pub fn is_prime(n: i32) -> bool {
    if n < 2 {
        return false;
    } else if n == 2 {
        return true;
    }
    let mut i = 2;
    while i * i < n {
        if n % i == 0 {
            return false;
        } else {
            i += 1;
            continue;
        }
    }
    true
}

pub fn lcm(mut n: i32, mut m: i32) -> i32 {
    n = if n > m { m } else { n };
    m = if n > m { n } else { m };
    let mut i = n;
    while i % m != 0 {
        i += n;
    }
    i
}

pub fn abs(n: i32) -> i32 {
    if n >= 0 {
        n
    } else {
        -n
    }
}

pub fn max(n: i32, m: i32) -> i32 {
    if n >= m {
        n
    } else {
        m
    }
}

pub fn min(n: i32, m: i32) -> i32 {
    if n <= m {
        n
    } else {
        m
    }
}

pub fn factorial(n: u32) -> u32 {
    if n == 0 {
        return 1
    }
    let mut r = n;
    for i in 0..n-1{
        r *= n-i-1;
    }
    r
}

pub fn combination(n: u32, m: u32) -> u32 {
    factorial(n) / factorial(m) / factorial(n-m)
}