Description

FlyBrother is a superman, therefore he is always busy saving the world.?
To graduate from NUDT is boring but necessary for him. Typically We need to post an paper to get Graduate Certificate, however being one superman, FlyBrother wants to make his paper perfect. A paper is a lower case string. To make it perfect, FlyBrother wanna the number of different substrings in the paper. It is quite a silly problem for FlyBrother, but because he is so busy, can you help him to solve it?

Input

There are several cases. Process till EOF.
For each case, there is a line of lower case string (1<=length<=100000).

Output

For each case, output one number in a line of the answer described in the problem.

Sample Input

a
aab

Sample Output

1
5

HINT

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;

#define LS 2*i
#define RS 2*i+1
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 100005
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define EXP 1e-8
LL wa[N],wb[N],wsf[N],wv[N],sa[N];
LL rank1[N],height[N],s[N],a[N];
char str[N],str1[N],str2[N];
//sa:字典序中排第i位的起始位置在str中第sa[i]
//rank:就是str第i個位置的後綴是在字典序排第幾
//height：字典序排i和i-1的後綴的最長公共前綴
LL cmp(LL *r,LL a,LL b,LL k)
{
    return r[a]==r[b]&&r[a+k]==r[b+k];
}
void getsa(LL *r,LL *sa,LL n,LL m)//n要包括末尾加入的0
{
    LL i,j,p,*x=wa,*y=wb,*t;
    for(i=0; i<m; i++)  wsf[i]=0;
    for(i=0; i<n; i++)  wsf[x[i]=r[i]]++;
    for(i=1; i<m; i++)  wsf[i]+=wsf[i-1];
    for(i=n-1; i>=0; i--)  sa[--wsf[x[i]]]=i;
    p=1;
    j=1;
    for(; p<n; j*=2,m=p)
    {
        for(p=0,i=n-j; i<n; i++)  y[p++]=i;
        for(i=0; i<n; i++)  if(sa[i]>=j)  y[p++]=sa[i]-j;
        for(i=0; i<n; i++)  wv[i]=x[y[i]];
        for(i=0; i<m; i++)  wsf[i]=0;
        for(i=0; i<n; i++)  wsf[wv[i]]++;
        for(i=1; i<m; i++)  wsf[i]+=wsf[i-1];
        for(i=n-1; i>=0; i--)  sa[--wsf[wv[i]]]=y[i];
        t=x;
        x=y;
        y=t;
        x[sa[0]]=0;
        for(p=1,i=1; i<n; i++)
            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)? p-1:p++;
    }
}
void getheight(LL *r,LL n)//n不保存最後的0
{
    LL i,j,k=0;
    for(i=1; i<=n; i++)  rank1[sa[i]]=i;
    for(i=0; i<n; i++)
    {
        if(k)
            k--;
        else
            k=0;
        j=sa[rank1[i]-1];
        while(r[i+k]==r[j+k])
            k++;
        height[rank1[i]]=k;
    }
}
LL t,ans,n,m;

int main()
{
    LL i,j,k,len;
    W(~scanf("%s",str))
    {
        len = strlen(str);
        UP(i,0,len-1)
        s[i]=str[i];
        s[len] = 0;
        getsa(s,sa,len+1,300);
        getheight(s,len);
        ans = (1+len)*len/2;
        UP(i,2,len)
        ans-=height[i];
        printf("%lld\n",ans);
    }
}

CSU1656: Paper of FlyBrother(後綴數組)