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

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 << 31) - 1);
const signed long long llf = (signed long long)((1ull << 61) - 1);
const double eps = 1e-6;
const int binary_limit = 128;
#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 = 1;
    while(!isdigit((x = getchar())) && x != '-' && x != -1);
    if(x == -1) {
        ungetc(x, stdin);    
        return false;
    }
    if(x == '-'){
        x = getchar();
        aFlag = -1;
    }
    for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0');
    ungetc(x, stdin);
    u *= aFlag;
    return true;
}

int n;
int m = 0;
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 = 0; i < n; i++) {
        static int c, x, l;
        cin >> strs[i];
        l = strs[i].length();
        readInteger(c);
        for(int j = 0; j < c; j++) {
            readInteger(x);
            pos[i].push_back(x);
            smax(m, x + l - 1);
        }
    }
}

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 + 2];
    suf = new int[m + 2];
    pre = new int[m + 2];
    res = new char[m + 2];
    memset(exist, true, sizeof(boolean) * (m + 2));
    suf[n + 1] = m + 1, pre[0] = 0;
    for(int i = 0; i <= m; i++)    suf[i] = i + 1;
    for(int i = 1; i <= m + 1; i++) pre[i] = i - 1;
    
    for(int i = 0; i < n; i++) {
        int l = strs[i].length();
        for(int j = 0; 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 = 1; i <= m; i++)
        if(exist[i])
            res[i] = 'a';
    res[m + 1] = 0;
    
    puts(res + 1);
}

int main() {
    init();
    solve();
    return 0;
}