Problem:
Nathan O. Davis has been running an electronic bulletin board system named JAG-channel. He is now having hard time to add a new feature there --- threaded view.

Like many other bulletin board systems, JAG-channel is thread-based. Here a thread (also called a topic) refers to a single conversation with a collection of posts. Each post can be an opening post, which initiates a new thread, or a reply to a previous post in an existing thread.

Threaded view is a tree-like view that reflects the logical reply structure among the posts: each post forms a node of the tree and contains its replies as its subnodes in the chronological order (i.e. older replies precede newer ones). Note that a post along with its direct and indirect replies forms a subtree as a whole.

Let us take an example. Suppose: a user made an opening post with a message hoge; another user replied to it with fuga; yet another user also replied to the opening post with piyo; someone else replied to the second post (i.e. fuga”) with foobar; and the fifth user replied to the same post with jagjag.
The tree of this thread would look like:

hoge
├─fuga
│　├─foobar
│　└─jagjag
└─piyo

For easier implementation, Nathan is thinking of a simpler format: the depth of each post from the opening post is represented by dots. Each reply gets one more dot than its parent post. The tree of the above thread would then look like:

hoge
.fuga
..foobar
..jagjag
.piyo

Your task in this problem is to help Nathan by writing a program that prints a tree in the Nathan's format for the given posts in a single thread.

Input

Input contains a single dataset in the following format:

n
k_1
M_1
k_2
M_2
:
:
k_n
M_n

The first line contains an integer n (1 ≤ n ≤ 1,000), which is the number of posts in the thread.
Then 2n lines follow.
Each post is represented by two lines:
the first line contains an integer k_i (k_1 = 0, 1 ≤ k_i < i for 2 ≤ i ≤ n) and indicates the i-th post is a reply to the k_i-th post;
the second line contains a string M_i and represents the message of the i-th post.
k_1 is always 0, which means the first post is not replying to any other post, i.e. it is an opening post.

Each message contains 1 to 50 characters, consisting of uppercase, lowercase, and numeric letters.

Output

Print the given n messages as specified in the problem statement.

Sample Input 1

1
0
icpc

Output for the Sample Input 1

icpc

Sample Input 2

5
0
hoge
1
fuga
1
piyo
2
foobar
2
jagjag

Output for the Sample Input 2

hoge
.fuga
..foobar
..jagjag
.piyo

Sample Input 3

8
0
jagjag
1
hogehoge
1
buhihi
2
fugafuga
4
ponyoponyo
5
evaeva
4
nowawa
5
pokemon

Output for the Sample Input 3

jagjag
.hogehoge
..fugafuga
...ponyoponyo
....evaeva
....pokemon
...nowawa
.buhihi

Sample Input 4

6
0
nakachan
1
fan
2
yamemasu
3
nennryou2
4
dannyaku4
5
kouzai11

Output for the Sample Input 4

nakachan
.fan
..yamemasu
...nennryou2
....dannyaku4
.....kouzai11

Sample Input 5

34
0
LoveLive
1
honoka
2
borarara
2
sunohare
2
mogyu
1
eri
6
kasikoi
7
kawaii
8
eriichika
1
kotori
10
WR
10
haetekurukotori
10
ichigo
1
umi
14
love
15
arrow
16
shoot
1
rin
18
nyanyanya
1
maki
20
6th
20
star
22
nishikino
1
nozomi
24
spiritual
25
power
1
hanayo
27
darekatasukete
28
chottomattete
1
niko
30
natsuiro
30
nikkonikkoni
30
sekaino
33
YAZAWA

Output for the Sample Input 5

LoveLive
.honoka
..borarara
..sunohare
..mogyu
.eri
..kasikoi
...kawaii
....eriichika
.kotori
..WR
..haetekurukotori
..ichigo
.umi
..love
...arrow
....shoot
.rin
..nyanyanya
.maki
..6th
..star
...nishikino
.nozomi
..spiritual
...power
.hanayo
..darekatasukete
...chottomattete
.niko
..natsuiro
..nikkonikkoni
..sekaino
...YAZAWA

Sample Input 6

6
0
2ch
1
1ostu
1
2get
1
1otsu
1
1ostu
3
pgr

Output for the Sample Input 6

2ch
.1ostu
.2get
..pgr
.1otsu
.1ostu

Solution in C++ (judge verdict: Accepted)

#define _USE_MATH_DEFINES
#define INF 0x3f3f3f3f
 
#include <iostream>
#include <cstdio>
#include <sstream>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <limits>
#include <map>
#include <string>
#include <cstring>
#include <set>
#include <deque>
#include <bitset>
#include <list>
#include <cctype>
#include <utility>
   
using namespace std;
   
typedef long long ll;
typedef pair <int,int> P;
typedef pair <int,P> PP;
   
static const double EPS = 1e-8;
   
static const int tx[] = {0,1,0,-1};
static const int ty[] = {-1,0,1,0};

bool visited[1001];

void dfs(int current,
         int depth,
         const vector<int> edges[1001],
         const vector<string>& messages){
  for(int i = 0; i < depth; i++){
    cout << ".";
  }
  cout << messages[current] << endl;
  for(int i = 0; i < edges[current].size(); i++){
    int next = edges[current][i];
    if(visited[next]) continue;
    visited[next] = true;
    dfs(next,depth+1,edges,messages);
  }
}

int main(){
  int num_of_posts;
  while(~scanf("%d",&num_of_posts)){
    memset(visited,false,sizeof(visited));

    vector<int> edges[1001];
    vector<string> messages;
    for(int post_i = 0; post_i < num_of_posts; post_i++){
      string message;
      int reply_to;
      cin >> reply_to >> message;
      messages.push_back(message);

      if(reply_to - 1 >= 0){
        edges[reply_to - 1].push_back(post_i);
      }
    }
    dfs(0,0,edges,messages);
  }
}