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 N = 1e6 + 10;
long long Ans;
int Siz[N << 1];
vector<int> Edge[N << 1];
struct SAM {
  int siz, lst;
  struct Node {
    int len, link, nxt[26];
  } t[N << 1];
  inline void Init() { siz = lst = 0, t[0].link = -1; }
  inline void Insert(int c) {
    int cur = ++siz, p = lst;
    t[cur].len = t[p].len + 1;
    while (~p && !t[p].nxt[c]) t[p].nxt[c] = cur, p = t[p].link;
    if (!~p)
      t[cur].link = 0;
    else {
      int q = t[p].nxt[c];
      if (t[p].len + 1 == t[q].len)
        t[cur].link = q;
      else {
        ++siz, t[siz] = t[q], t[siz].len = t[p].len + 1;
        while (~p && t[p].nxt[c] == q) t[p].nxt[c] = siz, p = t[p].link;
        t[q].link = t[cur].link = siz;
      }
    }
    Siz[lst = cur] = 1;
  }
} T;
char str[N];
void dfs(int now) {
  for (int v : Edge[now]) dfs(v), Siz[now] += Siz[v];
  Ans += 1ll * Siz[now] * (Siz[now] + 1) / 2 *
         (T.t[now].len - T.t[T.t[now].link].len);
}
inline void Work() {
  scanf("%s", str + 1);
  int n = strlen(str + 1);
  T.Init();
  for (register int i = 1; i <= n; ++i) T.Insert(str[i] - 'a');
  for (register int i = 1; i <= T.siz; ++i) Edge[T.t[i].link].push_back(i);
  dfs(0);
  cout << Ans << endl;
}
int main() { Work(); }