Problem: Eudokimus, a system administrator is in trouble again. As a result of an error in some script, a list of names of very important files has been damaged. Since they were files in the BerFS file system, it is known that each file name has a form "name.ext", where: 

  * name is a string consisting of lowercase Latin letters, its length is from 1 to 8 characters; 
  * ext is a string consisting of lowercase Latin letters, its length is from 1 to 3 characters. 



For example, "read.me", "example.txt" and "b.cpp" are valid file names and "version.info", "ntldr" and "contestdata.zip" are not.

Damage to the list meant that all the file names were recorded one after another, without any separators. So now Eudokimus has a single string.

Eudokimus needs to set everything right as soon as possible. He should divide the resulting string into parts so that each part would be a valid file name in BerFS. Since Eudokimus has already proved that he is not good at programming, help him. The resulting file list can contain the same file names.

Input

The input data consists of a single string s, its length is from 1 to 4·105 characters. The string can contain only lowercase Latin letters ('a' - 'z') and periods ('.').

Output

In the first line print "YES" (without the quotes), if it is possible to divide s into parts as required. In this case, the following lines should contain the parts of the required partition, one per line in the order in which they appear in s. The required partition can contain the same file names. If there are multiple solutions, print any of them.

If the solution does not exist, then print in a single line "NO" (without the quotes).

Examples

Input

read.meexample.txtb.cpp


Output

YES
read.m
eexample.t
xtb.cpp


Input

version.infontldrcontestdata.zip


Output

NO

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 1e1 + 4;
vector<int> circle;
ll vis[N];
bool bad_Relation[N];
int parent[N];
ll chk[N];
int color[N];
int tim[N];
ll dis[N];
int position[N];
vector<pair<ll, ll>> adj[N];
vector<int> adj1[N];
vector<int> graph[N];
bool has_cycle;
int maxdis, maxnode, Totnode, depth = 1;
bool ok;
queue<int> q;
stack<int> stk;
vector<int> solution;
int indegree[N];
int go[N];
int to[N];
ll x1, x2, x3, x4, x5, x6;
string ss;
int gn, gk, gcnt;
ll node, edge, u, v, cost;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  int cas = 0;
  while (t--) {
    ll n, m, i, j, q, cnt = 0, cnt1 = 0, cnt2 = 0, even = 0, odd = 0, len, k, r,
                      l, mid, z = 0, x = 0, y = 0, flag = 0, sum = 0,
                      tottal = 0;
    ll a = 0, b = 0, c = 0, d = 0, ans = 0LL, rem, quot, zero = 0, fst = 0,
       null = 0, snd = 0, lst = 0, rone = 0, one = 0, pos = 0, neg = 0,
       mn = INT_MAX, mx = INT_MIN;
    char ch;
    int h1, h2, m1, m2, h;
    int velo1, velo2, ac1, ac2, tim, hour, mint, sec;
    int bst, wrst;
    double nd, ad, bd, cd, dd, xd, sumd = 0.00;
    string str, str1 = "", str2 = "", str3 = "", strstore1 = "", strstore2 = "";
    cin >> str;
    vector<string> vec;
    i = 0;
    while (i < str.size()) {
      str1 = "";
      cnt = 0;
      while (str[i] != '.' && i < str.size()) {
        cnt++;
        str1 += str[i];
        i++;
      }
      if (cnt > 8 || cnt == 0 || i == str.size() || i + 1 == str.size()) {
        cout << "NO" << endl;
        return 0;
      }
      str1 += '.';
      i++;
      int nextpoint = -1;
      for (j = 0; j < 4 && nextpoint == -1 && i + j < str.size(); j++) {
        if (str[i + j] == '.') {
          nextpoint = j;
        }
      }
      if (nextpoint == 0 || nextpoint == 1) {
        cout << "NO" << endl;
        return 0;
      } else if (nextpoint != -1) {
        str1 += str[i++];
      } else {
        for (j = 0; j < 3 && i < str.size(); j++) {
          str1 += str[i++];
        }
      }
      vec.push_back(str1);
    }
    cout << "YES" << endl;
    for (auto it : vec) {
      cout << it << endl;
    }
  }
  return 0;
}