Codeforces Round #423 (Div. 2, rated, based on VK Cup Finals) Problem C (Codeforces 828C) - 链表 - 并查集

Ivan had string s consisting of small English letters. However, his friend Julia decided to make fun of him and hid the string s. Ivan preferred making a new string to finding the old one.

Ivan knows some information about the string s. Namely, he remembers, that string t_i occurs in string s at least k_i times or more, he also remembers exactly k_i positions where the string t_i occurs in string s: these positions are x_i, 1, x_i, 2, ..., x_i, ki. He remembers n such stringst_i.

You are to reconstruct lexicographically minimal string s such that it fits all the information Ivan remembers. Strings t_i and string sconsist of small English letters only.

Input

The first line contains single integer n (1 ≤ n ≤ 10⁵) — the number of strings Ivan remembers.

The next n lines contain information about the strings. The i-th of these lines contains non-empty string t_i, then positive integer k_i, which equal to the number of times the string t_i occurs in string s, and then k_i distinct positive integers x_i, 1, x_i, 2, ..., x_i, ki in increasing order — positions, in which occurrences of the string t_i in the string s start. It is guaranteed that the sum of lengths of strings t_i doesn't exceed10⁶, 1 ≤ x_i, j ≤ 10⁶, 1 ≤ k_i ≤ 10⁶, and the sum of all k_i doesn't exceed 10⁶. The strings t_i can coincide.

It is guaranteed that the input data is not self-contradictory, and thus at least one answer always exists.

Output

Print lexicographically minimal string that fits all the information Ivan remembers.

Examples

input

3
a 4 1 3 5 7
ab 2 1 5
ca 1 4

output

abacaba

input

1
a 1 3

output

aaa

input

3
ab 1 1
aba 1 3
ab 2 3 5

output

ababab

---

　　我同学都用排序 + 并查集，然而我用链表 + 并查集思想。

　　因为数据合法不用判错，所以用链表维护还未确定的位置。每条信息，找第一个确定的位置时边找边压缩路径(并查集思想)。然后按题意弄一下即可。

　　另外吐槽一下这道题。。直接交代码在第十个点各种WA，RE（必须交G++14才能AC，而且跑得非常慢），然后cf上交文件AC。。(可能是你在逗我)

Code

 /**
  * Codeforces
  * Problem#828C
  * Accepted
  * Time:296ms
  * Memory:59888k
  */
 #include <iostream>
 #include <cstdio>
 #include <ctime>
 #include <cmath>
 #include <cctype>
 #include <cstring>
 #include <cstdlib>
 #include <fstream>
 #include <sstream>
 #include <algorithm>
 #include <map>
 #include <set>
 #include <stack>
 #include <queue>
 #include <vector>
 #include <stack>
 #ifndef WIN32
 #define Auto "%lld"
 #else
 #define Auto "%I64d"
 #endif
 using namespace std;
 typedef bool boolean;
 const signed int inf = (signed)((1u << ) - );
 const signed long long llf = (signed long long)((1ull << ) - );
 const double eps = 1e-;
 const int binary_limit = ;
 #define smin(a, b) a = min(a, b)
 #define smax(a, b) a = max(a, b)
 #define max3(a, b, c) max(a, max(b, c))
 #define min3(a, b, c) min(a, min(b, c))
 template<typename T>
 inline boolean readInteger(T& u){
     char x;
     int aFlag = ;
     while(!isdigit((x = getchar())) && x != '-' && x != -);
     if(x == -) {
         ungetc(x, stdin);
         return false;
     }
     if(x == '-'){
         x = getchar();
         aFlag = -;
     }
     for(u = x - ''; isdigit((x = getchar())); u = (u << ) + (u << ) + x - '');
     ungetc(x, stdin);
     u *= aFlag;
     return true;
 }

 int n;
 int m = ;
 string* strs;
 vector<int> *pos;
 boolean* exist;
 char* res;
 int* suf;
 int* pre;

 inline void init() {
     readInteger(n);
     strs = new string[n];
     pos = new vector<int>[n];
     for(int i = ; i < n; i++) {
         static int c, x, l;
         cin >> strs[i];
         l = strs[i].length();
         readInteger(c);
         for(int j = ; j < c; j++) {
             readInteger(x);
             pos[i].push_back(x);
             smax(m, x + l - );
         }
     }
 }

 inline void remove(int x) {
     pre[suf[x]] = pre[x];
     suf[pre[x]] = suf[x];
     exist[x] = false;
 }

 int find(int x, int endpos) {
     if(exist[x] || x >= endpos)    return x;
     return suf[x] = find(suf[x], endpos);
 }

 inline void solve() {
     exist = new boolean[m + ];
     suf = new int[m + ];
     pre = new int[m + ];
     res = new char[m + ];
     memset(exist, true, sizeof(boolean) * (m + ));
     suf[n + ] = m + , pre[] = ;
     for(int i = ; i <= m; i++)    suf[i] = i + ;
     for(int i = ; i <= m + ; i++) pre[i] = i - ;

     for(int i = ; i < n; i++) {
         int l = strs[i].length();
         for(int j = ; j < (signed)pos[i].size(); j++) {
             int p = pos[i][j], start = pos[i][j];
             p = find(p, start + l);
             while(p < pos[i][j] + l) {
                 res[p] = strs[i][p - start];
                 remove(p);
                 p = suf[p];
             }
         }
     }

     for(int i = ; i <= m; i++)
         if(exist[i])
             res[i] = 'a';
     res[m + ] = ;

     puts(res + );
 }

 int main() {
     init();
     solve();
     return ;
 }