字符串（后缀自动机）：Codeforces Round #129 (Div. 1) E.Little Elephant and Strings

E. Little Elephant and Strings

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The Little Elephant loves strings very much.

He has an array a from n strings, consisting of lowercase English letters. Let's number the elements of the array from 1 to n, then let's denote the element number i as a_i. For each string a_i (1 ≤ i ≤ n) the Little Elephant wants to find the number of pairs of integers l and r (1 ≤ l ≤ r ≤ |a_i|) such that substring a_i[l... r] is a substring to at least k strings from array a (including the i-th string).

Help the Little Elephant solve this problem.

If you are not familiar with the basic notation in string problems, you can find the corresponding definitions in the notes.

Input

The first line contains two space-separated integers — n and k (1 ≤ n, k ≤ 10⁵). Next n lines contain array a. The i-th line contains a non-empty string a_i, consisting of lowercase English letter. The total length of all strings a_i does not exceed 10⁵.

Output

On a single line print n space-separated integers — the i-th number is the answer for string a_i.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

3 1
abc
a
ab

Output

6 1 3 

Input

7 4
rubik
furik
abab
baba
aaabbbababa
abababababa
zero

Output

1 0 9 9 21 30 0 

Note

Let's assume that you are given string a = a₁a₂... a_|a|, then let's denote the string's length as |a| and the string's i-th character as a_i.

A substring a[l... r] (1 ≤ l ≤ r ≤ |a|) of string a is string a_la_l + 1... a_r.

String a is a substring of string b, if there exists such pair of integers l and r (1 ≤ l ≤ r ≤ |b|), that b[l... r] = a.

　　这道题很好啊……

　　本来要用后缀数组，现在我用的是后缀自动机。用set维护经过的字串有哪些，建成这棵trie的SAM后，连边fa[i]->i，然后DFS，把set启发式合并一下，要的信息就是size，处理答案时先跳（必定可行），然后size<k就不停跳fa，直到size>=k为止，这时匹配的答案加入计数。

#include <iostream>
#include <cstring>
#include <cstring>
#include <cstdio>
#include <set>
using namespace std;
const int N=200010;
int fa[N],ch[N][27],len[N];
string s[N];set<int>t[N];
int tr[N][26],last[N];
int lst,cnt,n,k,tot;
void Insert(int c){
    int p=lst,np=lst=++cnt;len[np]=len[p]+1;
    while(p&&ch[p][c]==0)ch[p][c]=np,p=fa[p];
    if(!p)fa[np]=1;
    else{
        int q=ch[p][c],nq;
        if(len[q]==len[p]+1)
            fa[np]=q;
        else{
            len[nq=++cnt]=len[p]+1;
            fa[nq]=fa[q];fa[q]=fa[np]=nq;
            memcpy(ch[nq],ch[q],sizeof(ch[q]));
            while(ch[p][c]==q)ch[p][c]=nq,p=fa[p];
        }    
    }
}
int cntE,fir[N],to[N],nxt[N];
void addedge(int a,int b){
    nxt[++cntE]=fir[a];
    to[fir[a]=cntE]=b;
}
void Initialization(){
    memset(fir,0,sizeof(fir));
    lst=cnt=1;last[cntE=tot=0]=1;
}
set<int>Merge(set<int>a,set<int>b){
    if(a.size()<b.size())swap(a,b);
    a.insert(b.begin(),b.end());
    delete &b;return a;
}
void DFS(int x){
    for(int i=fir[x],g;i;i=nxt[i])
        DFS(g=to[i]),t[x]=Merge(t[x],t[g]);
}

int main(){
    Initialization();
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++){
        cin>>s[i];
        for(int j=0,p=0,c;j<s[i].size();j++){
            c=s[i][j]-'a';lst=last[p];
            if(tr[p][c])p=tr[p][c],t[last[p]].insert(i);
            else Insert(c),t[last[p=tr[p][c]=++tot]=lst].insert(i);
        }
    }
    for(int i=2;i<=cnt;i++)
        addedge(fa[i],i);DFS(1);
    for(int i=1;i<=n;i++){
        long long ans=0;
        for(int j=0,p=1,c,l=0;j<s[i].size();j++){
            c=s[i][j]-'a';
            p=ch[p][c];l+=1;
            while(p!=1&&t[p].size()<k)
                p=fa[p],l=len[p];
            ans+=l;    
        }
        printf("%I64d ",ans);
    }
    printf("
");
    return 0;
}