CodeForces

String Reconstruction

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_idoesn'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


寻找一个字符串。给你一些线索（某些子串的位置），输出一个满足条件的最小字典序字符串。
暴力求解超时。可以采用并查集，若当前位置已经有子串放置，那就从已知子串的末尾继续放。并查集记录下末尾的下一位。很好的避免了重叠部分的重复枚举。有点像cys学长教过的那道next跳的题。
并查集套路题。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<string>
#include<math.h>
#include<queue>
#include<set>
#include<stack>
#include<algorithm>
#include<vector>
#include<iterator>
#define MAX 2000005
#define INF 0x3f3f3f3f
#define MOD 1000000007
using namespace std;

typedef long long ll;

int f[MAX];
char a[MAX];
int find(int x){
    return f[x]==x?x:f[x]=find(f[x]);
}
int main()
{
    int n,m,x,len,i,j,k,ii;
    char s[MAX];
    scanf("%d",&n);
    for(i=0;i<=2000000;i++){
        a[i]='a';
        f[i]=i;
    }
    int maxx=0;
    while(n--){
        scanf(" %s %d",s,&k);
        len=strlen(s);
        for(j=1;j<=k;j++){
            scanf("%d",&x);
            int xx=find(x);
            int ff=find(len+x);
            for(i=xx,ii=xx-x;i<=len+x-1;i++,ii++){
                a[i]=s[ii];
                f[find(i)]=ff;
            }
            if(len+x-1>maxx) maxx=len+x-1;
        }
    }
    for(i=1;i<=maxx;i++){
        printf("%c",a[i]);
    }
    printf("
");
    return 0;
}