Given a string, we need to find the total number of its distinct substrings.

Input

T- number of test cases. T<=20; Each test case consists of one string, whose length is <= 50000

Output

For each test case output one number saying the number of distinct substrings.

Example

Input:
2
CCCCC
ABABA

Output:
5
9

【分析】
嗯…參考 論文

…nei個神奇的spoj交不了題…所以現在暫時未通過…我把樣例過了假裝自己A了吧

【程式碼】

#include<iostream>
#include<cstring>
#include<cstdio>
#define ll long long
#define M(a) memset(a,0,sizeof a)
#define fo(i,j,k) for(i=j;i<=k;i++)
using namespace std;
const int mxn=100005;
int a[mxn],b[mxn],x[mxn],y[mxn],sa[mxn],rank[mxn],height[mxn];
char
 
 s[mxn];
int n,m,len,T;
inline bool comp(int i,int j,int l)
{
    return y[i]==y[j]&&(i+l>len?-1:y[i+l])==(j+l>len?-1:y[j+l]);
}
inline void work()
{
    int i,j,k,p;
    fo(i,0,m) b[i]=0;
    fo(i,1,len) b[x[i]=a[i]]++;
    fo(i,1,m) b[i]+=b[i-1];
    for(i=len;i>=1;i--) sa[b[x[i]]--]=i;
    for
 
(k=1;k<=len;k<<=1)
    {
        p=0;   //處理第二關鍵字 
        fo(i,len-k+1,len) y[++p]=i;
        fo(i,1,len) if(sa[i]>k) y[++p]=sa[i]-k;
        //處理第一關鍵字 
        fo(i,0,m) b[i]=0;
        fo(i,1,len) b[x[y[i]]]++;
        fo(i,1,m) b[i]+=b[i-1];
        for(i=len;i>=1;i--) sa[b[x[y[i]]]--]=y[i];
        swap(x,y),p=2,x[sa[1]]=1;
        fo(i,2,len)
          x[sa[i]]=comp(sa[i-1],sa[i],k)?p-1:p++;
        if(p>len) break;
        m=p;
    }
    p=k=0;
    fo(i,1,len) rank[sa[i]]=i;
    for(i=1;i<=len;height[rank[i++]]=k)
      for(k?k--:0,j=sa[rank[i]-1];a[i+k]==a[j+k];k++);
}
int main()
{
    int i,j,ans;
    scanf("%d",&T);
    while(T--)
    {
        M(a);M(sa);
        ans=0;
        m=129;
        scanf("%s",s+1);
        len=strlen(s+1);
        fo(i,1,len) a[i]=s[i]; 
        work();
        fo(i,1,len) ans+=len-sa[i]+1-height[i];
        printf("%d\n",ans);
    }
    return 0;
}
/*
2 
CCCCC 
*/