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>
const int N = 1e5 + 10;
using ll = long long;
struct sam {
  static const int N = 2e5 + 10;
  int tot, lst, tr[N][26], Max[N], par[N], siz[N];
  inline sam() { tot = lst = 1; }
  inline void ins(int ch) {
    int Now = ++tot, p = lst;
    lst = Now, Max[Now] = Max[p] + 1, siz[Now] = 1;
    for (; p && !tr[p][ch]; p = par[p]) tr[p][ch] = Now;
    if (!p) return par[Now] = 1, void();
    int q = tr[p][ch];
    if (Max[q] == Max[p] + 1) return par[Now] = q, void();
    int New = ++tot;
    Max[New] = Max[p] + 1, par[New] = par[q];
    for (int i = 0; i < 26; ++i) tr[New][i] = tr[q][i];
    for (; p && tr[p][ch] == q; p = par[p]) tr[p][ch] = New;
    par[q] = par[Now] = New;
  }
  int tax[N], a[N];
  inline void solve() {
    ll ans = 0;
    for (int i = 1; i <= tot; ++i) tax[Max[i]]++;
    for (int i = 1; i <= tot; ++i) tax[i] += tax[i - 1];
    for (int i = 1; i <= tot; ++i) a[tax[Max[i]]--] = i;
    for (int i = tot; i >= 1; --i) siz[par[a[i]]] += siz[a[i]];
    for (int i = 1; i <= tot; ++i)
      ans += 1LL * siz[i] * (siz[i] + 1) / 2 * (Max[i] - Max[par[i]]);
    printf("%lld\n", ans);
  }
} T;
int main() {
  std::ios::sync_with_stdio(false), std::cin.tie(0);
  std::string str;
  std::cin >> str;
  for (int i = 0, len = str.length(); i < len; ++i) T.ins(str[i] - 'a');
  return T.solve(), 0;
}