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 Go (1.14.1) (judge verdict: Accepted)

package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

var sc = bufio.NewScanner(os.Stdin)

func Scan() string {
	sc.Scan()
	return sc.Text()
}

func iScan() int {
	n, _ := strconv.Atoi(Scan())
	return n
}

func fScan() float64 {
	n, _ := strconv.ParseFloat(Scan(), 64)
	return n
}
func stringToInt(s string) int {
	n, _ := strconv.Atoi(s)
	return n
}
func SScan(n int) []string {
	a := make([]string, n)
	for i := 0; i < n; i++ {
		a[i] = Scan()
	}
	return a
}
func iSScan(n int) []int {
	a := make([]int, n)
	for i := 0; i < n; i++ {
		a[i] = iScan()
	}
	return a
}
func abs(x int) int {
	if x < 0 {
		x = -x
	}
	return x
}
func larger(a, b int) int {
	if a < b {
		return b
	} else {
		return a
	}
}
func smaller(a, b int) int {
	if a > b {
		return b
	} else {
		return a
	}
}
func max(a []int) int {
	x := a[0]
	for i := 0; i < len(a); i++ {
		if x < a[i] {
			x = a[i]
		}
	}
	return x
}
func min(a []int) int {
	x := a[0]
	for i := 0; i < len(a); i++ {
		if x > a[i] {
			x = a[i]
		}
	}
	return x
}
func sum(a []int) int {
	x := 0
	for _, v := range a {
		x += v
	}
	return x
}

var mod int = 1000000007

func main() {
	buf := make([]byte, 0)
	sc.Buffer(buf, mod)
	sc.Split(bufio.ScanWords)
	s := Scan()
	fmt.Print(string(s[0]))
	fmt.Print(len(s) - 2)
	fmt.Println(string(s[len(s)-1]))
}