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;
struct SAM {
  SAM(void) {
    lst = tot = 0;
    memset(t->ch, 0xff, 26 << 2);
    t->fa = -1, t->len = 0;
  }
  struct SAMnode {
    int ch[26], fa, len;
  } t[(100005) << 1];
  int lst, tot, ind[(100005) << 1], q[(100005) << 1], Head, Tail,
      siz[(100005) << 1];
  void append(int c) {
    int p = lst, np = lst = ++tot;
    siz[np] = 1;
    memset(t[np].ch, 0xff, 26 << 2);
    t[np].len = t[p].len + 1;
    while (~p && !~t[p].ch[c]) t[p].ch[c] = np, p = t[p].fa;
    if (~p) {
      int q = t[p].ch[c];
      if (t[p].len + 1 == t[q].len) {
        t[np].fa = q;
      } else {
        int nq = ++tot;
        t[nq] = t[q];
        t[nq].len = t[p].len + 1;
        t[q].fa = t[np].fa = nq;
        while (~p && t[p].ch[c] == q) t[p].ch[c] = nq, p = t[p].fa;
      }
    } else {
      t[np].fa = 0;
    }
  }
  void solve(void) {
    memset(ind, 0, (tot + 1) << 2);
    for (int i = 1; i <= tot; ++i) ++ind[t[i].fa];
    Head = Tail = 0;
    for (int i = 1; i <= tot; ++i)
      if (!ind[i]) q[Tail++] = i;
    long long res(0);
    while (Head < Tail) {
      int x = q[Head++];
      if (x) {
        siz[t[x].fa] += siz[x];
        if (!--ind[t[x].fa]) q[Tail++] = t[x].fa;
        res += (long long)(t[x].len - t[t[x].fa].len) *
               ((long long)siz[x] * (siz[x] + 1) >> 1);
      }
    }
    cout << res << endl;
  }
} sam;
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  char c;
  while (cin >> c) sam.append(c - 'a');
  sam.solve();
  return 0;
}