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 k = 26;
struct node {
  node *to[k] = {0};
  node *link = {0};
  int len = 0;
  long long cnt = 0;
};
int main() {
  string s;
  cin >> s;
  node *root = new node;
  node *cur = root;
  vector<node *> a;
  for (char _c : s) {
    int c = (int)(_c - 'a');
    node *v = cur;
    cur = new node;
    a.push_back(cur);
    cur->cnt = 1;
    cur->len = v->len + 1;
    while (v && !v->to[c]) v->to[c] = cur, v = v->link;
    if (!v)
      cur->link = root;
    else if (v->len + 1 == v->to[c]->len)
      cur->link = v->to[c];
    else {
      node *u = v->to[c], *clone = new node;
      memcpy(clone->to, u->to, sizeof(u->to));
      clone->link = u->link;
      clone->len = v->len + 1;
      while (v && v->to[c] == u) v->to[c] = clone, v = v->link;
      cur->link = u->link = clone;
      a.push_back(clone);
    }
  }
  sort(a.begin(), a.end(), [](node *v, node *u) { return v->len > u->len; });
  long long ans = 0;
  for (node *v : a) {
    int len = v->len - v->link->len;
    ans += len * v->cnt * (v->cnt + 1) / 2;
    v->link->cnt += v->cnt;
  }
  cout << ans;
  return 0;
}