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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <utility>
#include <map>
#include <list>
#include <climits>
#include <set>
#include <deque>
#include <unordered_map>
#include <unordered_set>
#include <iomanip>
#include <bitset>

#define DEBUG(x) cout << "> " << #x << ": " << x << "\n";

using namespace std;

template <typename T>
std::ostream &operator <<(std::ostream &out, vector<T> &v) {
	for (typename vector<T>::size_type i = 0; i < v.size(); ++i)
		out << v[i] << " ";
	out << "\n";
    return out;
}
template <typename T>
std::ostream &operator <<(std::ostream &out, vector<pair<T, T> > &v) {
	for (size_t i = 0; i < v.size(); ++i)
		out << "(" << v[i].first << ", " << v[i].second << ") ";
	out << "\n";
    return out;
}
template <typename T>
std::ostream &operator <<(std::ostream &out, vector<vector<T> > &v) {
	for (size_t i = 0; i < v.size(); ++i) {
		for (size_t j = 0; j < v[i].size(); ++j) {
			out << v[i][j] << " ";
		}
		out << "\n";
	}
   	return out;
}

int main() {

	std::ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	string s;
	cin >> s;

	int sz = s.size();
	cout << s[0] << sz - 2 << s[sz - 1] << "\n";

	return 0;

}