CodeForces 547E：Mike and Friends（AC自动机+DFS序+主席树）

What-The-Fatherland is a strange country! All phone numbers there are strings consisting of lowercase English letters. What is double strange that a phone number can be associated with several bears!

In that country there is a rock band called CF consisting of n bears (including Mike) numbered from 1 to n.

Phone number of i-th member of CF is s_i. May 17th is a holiday named Phone Calls day. In the last Phone Calls day, everyone called all the numbers that are substrings of his/her number (one may call some number several times). In particular, everyone called himself (that was really strange country).

Denote as call(i, j) the number of times that i-th member of CF called the j-th member of CF.

The geek Mike has q questions that he wants to ask you. In each question he gives you numbers l, r and k and you should tell him the number

Input

The first line of input contains integers n and q (1 ≤ n ≤ 2 × 10⁵ and 1 ≤ q ≤ 5 × 10⁵).

The next n lines contain the phone numbers, i-th line contains a string s_iconsisting of lowercase English letters ().

The next q lines contain the information about the questions, each of them contains integers l, r and k (1 ≤ l ≤ r ≤ n and 1 ≤ k ≤ n).

Output

Print the answer for each question in a separate line.

Examples

Input

5 5
a
ab
abab
ababab
b
1 5 1
3 5 1
1 5 2
1 5 3
1 4 5

Output

7
5
6
3
6

题意：给定N个串，M次询问，每次询问[L,R]里含多少个X。

思路：（第一次做，结合主席树那里还是不太好想）。对N个串建立AC自动机，建立Fail树，然后得到DFS序。那么，对于每个串S，假设其长度为L，S在AC自动机上面跑，其每个前缀Si在AC自动机上面得到最大深度，对应的Fail树位置，贡献加1，保证这个贡献在Si的后缀的子树里（比如abcdef，那么跑到abcd时，在fail树上面对应的位置贡献加1 ，对于后缀abcd,bcd,cd,d的子树都含这个贡献）。 

        关键是如何出现子树来自于[L,R]的贡献。开始我以为以dfs序为X轴，以来自的串为Y轴建立主席树，查询的时候查询区间[in[u],out[u]]，即子树里关键字在[L,R]里的个数。但是发现没有转移关系。所以想不走了。

看了其他人的代码，是以每个串，以来自的串为X轴（从长的前缀到短的前缀），以dfs序为X轴，保证了前缀和的正确性，查询的时候查询区间[L-1,R]的区间在[in[[u],out[u]]的数量。

#include<bits/stdc++.h>
using namespace std;
const int maxn=200010;
int ch[maxn][26],fa[maxn],cnt=0; //trie树 
int Laxt[maxn],Next[maxn],To[maxn],tot; //fail树 
int q[maxn],fail[maxn],head,tail; //fail树 
int in[maxn],out[maxn],pos[maxn],times;//dfs序 
int p[maxn],rt[maxn],cur,num; struct in{int l,r,sum;}s[maxn*20];
char c[maxn];
void addedge(int u,int v){ Next[++tot]=Laxt[u]; Laxt[u]=tot; To[tot]=v; }
int insert()
{
    int Now=0; for(int i=1;c[i];i++){
        if(!ch[Now][c[i]-'a']){ 
             ch[Now][c[i]-'a']=++cnt;
             fa[cnt]=Now;
        }
        Now=ch[Now][c[i]-'a'];
    }   return Now;
}
void buildfail()
{
    for(int i=0;i<26;i++){
        if(ch[0][i]) q[++head]=ch[0][i],fail[ch[0][i]]=0;
        else ch[0][i]=0;
    }
    while(tail<head){
        int Now=q[++tail];
        for(int i=0;i<26;i++){
            if(ch[Now][i]) {
                q[++head]=ch[Now][i]; fail[ch[Now][i]]=ch[fail[Now]][i];
            }
            else ch[Now][i]=ch[fail[Now]][i];
        }
    }
    for(int i=1;i<=cnt;i++) addedge(fail[i],i);
}
void dfs(int u)
{
    in[u]=++times;
    for(int i=Laxt[u];i;i=Next[i]) dfs(To[i]);
    out[u]=times;
}
void add(int &Now,int pre,int L,int R,int pos)
{
    Now=++num; s[Now]=s[pre]; s[Now].sum++;
    if(L==R) return ; int Mid=(L+R)>>1;
    if(pos<=Mid) add(s[Now].l,s[pre].l,L,Mid,pos);
    else add(s[Now].r,s[pre].r,Mid+1,R,pos);
}
int query(int pre,int Now,int L,int R,int l,int r)
{
    if(l<=L&&r>=R) return s[Now].sum-s[pre].sum;
    int res=0,Mid=(L+R)>>1;
    if(l<=Mid) res+=query(s[pre].l,s[Now].l,L,Mid,l,r);
    if(r>Mid) res+=query(s[pre].r,s[Now].r,Mid+1,R,l,r);
    return res;
}
int main()
{
    int N,M,Q,L,R,x,i,j;
    scanf("%d%d",&N,&Q);
    for(i=1;i<=N;i++){
        scanf("%s",c+1);
        pos[i]=insert();
    }
    buildfail();
    dfs(0); cur=0;
    for(i=1;i<=N;i++){
        for(j=pos[i];j;j=fa[j]){
            cur++;
            add(rt[cur],rt[cur-1],1,times,in[j]);
        }
        p[i]=rt[cur];
    }
    while(Q--){
        scanf("%d%d%d",&L,&R,&x);
        printf("%d
",query(p[L-1],p[R],1,times,in[pos[x]],out[pos[x]]));
    }
    return 0;
}