Problem: You are given a string s. Each pair of numbers l and r that fulfill the condition 1 ≤ l ≤ r ≤ |s|, correspond to a substring of the string s, starting in the position l and ending in the position r (inclusive).

Let's define the function of two strings F(x, y) like this. We'll find a list of such pairs of numbers for which the corresponding substrings of string x are equal to string y. Let's sort this list of pairs according to the pair's first number's increasing. The value of function F(x, y) equals the number of non-empty continuous sequences in the list.

For example: F(babbabbababbab, babb) = 6. The list of pairs is as follows:

(1, 4), (4, 7), (9, 12)

Its continuous sequences are: 

  * (1, 4)
  * (4, 7)
  * (9, 12)
  * (1, 4), (4, 7)
  * (4, 7), (9, 12)
  * (1, 4), (4, 7), (9, 12)



Your task is to calculate for the given string s the sum F(s, x) for all x, that x belongs to the set of all substrings of a string s.

Input

The only line contains the given string s, consisting only of small Latin letters (1 ≤ |s| ≤ 105).

Output

Print the single number — the sought sum.

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

Examples

Input

aaaa


Output

20


Input

abcdef


Output

21


Input

abacabadabacaba


Output

188

Note

In the first sample the function values at x equal to "a", "aa", "aaa" and "aaaa" equal 10, 6, 3 and 1 correspondingly.

In the second sample for any satisfying x the function value is 1.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int Max_N = 1e5;
int n, cnt = 1, lst = 1, Num, lin[Max_N * 2 + 5], T[Max_N * 2 + 5][26],
       len[Max_N * 2 + 5], p[Max_N * 2 + 5], tot[Max_N * 2 + 5];
char s[Max_N + 5];
long long ans;
struct kk {
  int Id, Next;
} e[Max_N * 2 + 5];
inline void Insert(int x, int y) {
  e[++Num].Next = lin[x];
  lin[x] = Num;
  e[Num].Id = y;
}
inline void Add(int x) {
  int New = ++cnt;
  len[New] = len[lst] + 1;
  tot[New] = 1;
  for (; lst && !T[lst][x]; T[lst][x] = New, lst = p[lst])
    ;
  if (!lst)
    p[New] = 1;
  else {
    int q = T[lst][x];
    if (len[lst] + 1 == len[q])
      p[New] = q;
    else {
      int nq = ++cnt;
      p[nq] = p[q];
      p[q] = p[New] = nq;
      len[nq] = len[lst] + 1;
      for (int i = 0; i < 26; i++) T[nq][i] = T[q][i];
      for (; lst && T[lst][x] == q; T[lst][x] = nq, lst = p[lst])
        ;
    }
  }
  lst = New;
}
inline void dfs(int x) {
  for (int i = lin[x]; i; i = e[i].Next) {
    dfs(e[i].Id);
    tot[x] += tot[e[i].Id];
  }
  ans += 1ll * tot[x] * (tot[x] + 1) / 2 * (len[x] - len[p[x]]);
}
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (int i = 1; i <= n; i++) Add(s[i] - 'a');
  for (int i = 1; i <= cnt; i++) Insert(p[i], i);
  dfs(1);
  cout << ans << endl;
  return 0;
}