CodeForces - 827A：String Reconstruction （基础并查集）

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_ioccurs 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 strings t_i.

You are to reconstruct lexicographically minimal string s such that it fits all the information Ivan remembers. Strings t_i and string s consist 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 exceed 10⁶, 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

题意：给定一个字符串的部分消息，让你将其补齐，并且字典序最小。

思路：给出的消息可以得到字符串长度。现在就说需要把给出的消息都填进去，没有填到的部分用‘a’补齐，给出的消息很多重复的，直接for是不想的，我们用并查集快速的得到后面的第一个空着的位置。

（常规操作，居然想不到。。。

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=2000010;
char c[maxn]; int ans[maxn],fa[maxn];
int find(int x){
    if(x!=fa[x]) fa[x]=find(fa[x]);
    return fa[x];
}
int main()
{
    int N,L=0;
    scanf("%d",&N);
    rep(i,1,2e6) fa[i]=i;
    rep(i,1,N){
        scanf("%s",c+1);
        int Len=strlen(c+1),num; scanf("%d",&num);
        rep(j,1,num){
            int x,pos; scanf("%d",&x); L=max(L,x+Len-1);
            pos=x; while(true){
                pos=find(pos);
                if(pos>x+Len-1||!pos) break;
                ans[pos]=c[pos-x+1];
                fa[pos]=fa[find(pos+1)];
                pos=fa[pos];
            }
        }
    }
    rep(i,1,L){
        if(ans[i]=='\0') ans[i]='a';
        putchar(ans[i]);
    }
    return 0;
}