//Trie树（字典树）练习 
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <set>
#include <queue>
using namespace std;

const int MAXN = 1000010;
char str[MAXN], Ans[MAXN];
int exist[MAXN], End[MAXN], len, Bound[MAXN];
int eid, cnt, nex[MAXN][30], Type[MAXN], type[MAXN], Eid;
set<int> G[MAXN][2];
bool E[MAXN][2];
// 空格 0, 句号 1，逗号 2

void dfs(int k, int t) {
	//cout << k << " " << t << endl; 
	for (auto y: G[k][t]) {
		if (!E[y][1 - t]) {
			E[y][1 - t] = true;
			dfs(y, 1 - t);
		}
	}
}

int main() {
	//freopen("test.in", "r", stdin);
	//freopen("test.out", "w", stdout);
	for (int i = 1; i < MAXN; i++) exist[i] = -1; 
	while (scanf("%s", str + 1) != EOF) {
		int n = strlen(str + 1), m;
		cnt++;
		if ('a' <= str[n] && str[n] <= 'z') m = n;
		else m = n - 1;
		Type[cnt] = (Type[cnt - 1] % 3) * 3;
		if ('.' == str[n]) Type[cnt] ++;
		if (',' == str[n]) Type[cnt] += 2; 
		for (int i = len + 1; i <= len + m; i++)
			Ans[i] = str[i - len];
		len = len + m;
		Bound[cnt] = len;
		int p = 0;
		for (int i = 1; i <= m; i++) {
			int c = str[i] - 'a';
			if (!nex[p][c]) nex[p][c] = ++eid;
			p = nex[p][c];
 		}
 		if (exist[p] == -1) {
 			type[p] = ++Eid;
 			//printf("%d %s\n", Eid, str + 1);
		}
 		if (exist[p] == -1)
		 	exist[p] = 0;
 		if (Type[cnt] % 3 == 2 && exist[p] % 3 == 0)
 			exist[p] += 2;
 		if (Type[cnt] / 3 == 2 && exist[p] / 3 == 0)
 			exist[p] += 6; 
 		End[cnt] = p;
	}
	for (int i = 1; i <= cnt; i++) {
		if (Type[i + 1] / 3 != 1) {
			//cout << "Yes! " << type[End[i]] << " " << type[End[i + 1]] << endl;
			G[type[End[i]]][0].insert(type[End[i + 1]]); // ..., -> ,... 
		}
		if (Type[i - 1] % 3 != 1 && !E[type[End[i - 1]]][1]) {
			//cout << "Yes2! " << type[End[i]] << " " << type[End[i - 1]] << endl;
			G[type[End[i]]][1].insert(type[End[i - 1]]); // ,... -> ...,
		}
	} 
	for (int i = 1; i <= Eid; i++) E[i][0] = E[i][1] = false;
	for (int i = 1; i <= cnt; i++) {
		if (exist[End[i]] % 3 == 2 && !E[type[End[i]]][0]) {
			E[type[End[i]]][0] = true;
			dfs(type[End[i]], 0);
		}
		if (exist[End[i]] / 3 == 2 && !E[type[End[i]]][1]) {
			E[type[End[i]]][1] = true;
			dfs(type[End[i]], 1);
		} 
	} 
//	for (int i = 1; i <= Eid; i++)
//		cout << E[i][0] << " " << E[i][1] << endl;
	for (int i = 1; i <= cnt; i++) {
		for (int j = Bound[i - 1] + 1; j <= Bound[i]; j++)
			printf("%c", Ans[j]); 
		if (Type[i] % 3 != 1 && (E[type[End[i]]][0] || E[type[End[i + 1]]][1])) printf(",");
		if (Type[i] % 3 == 1) printf(".");
		printf(" ");
	}
	//fclose(stdin);
	//fclose(stdout);
	return 0;
}