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 Node {
  int len, link;
  bool term;
  map<char, int> go;
  Node() : len(0), link(-1), term(false) {}
};
int n;
Node sa[200050];
int sa_size = 1, last;
int cnt[200050];
void extend(char c) {
  int cur = sa_size++;
  sa[cur].len = sa[last].len + 1;
  int p = last;
  for (; p != -1 && sa[p].go.count(c) == 0; p = sa[p].link) sa[p].go[c] = cur;
  if (p == -1) {
    sa[cur].link = 0;
  } else {
    int q = sa[p].go[c];
    if (sa[q].len == sa[p].len + 1) {
      sa[cur].link = q;
    } else {
      int clone = sa_size++;
      sa[clone] = sa[q];
      sa[clone].len = sa[p].len + 1;
      sa[q].link = clone;
      for (; p != -1 && sa[p].go[c] == q; p = sa[p].link) sa[p].go[c] = clone;
      sa[cur].link = clone;
    }
  }
  last = cur;
}
void mark_terminal() {
  for (int p = last; p != -1; p = sa[p].link) sa[p].term = true;
}
int get_cnt(int v) {
  if (cnt[v] == -1) {
    cnt[v] = (sa[v].term ? 1 : 0);
    for (auto it = sa[v].go.begin(); it != sa[v].go.end(); it++) {
      int to = it->second;
      cnt[v] += get_cnt(to);
    }
  }
  return cnt[v];
}
int main() {
  string str;
  cin >> str;
  for (int i = 0; i < (int)str.size(); i++) {
    extend(str[i]);
  }
  memset(cnt, -1, sizeof(cnt));
  mark_terminal();
  long long sum = 0;
  for (int i = 1; i < sa_size; i++) {
    long long val = get_cnt(i);
    sum += val * (val + 1) / 2 * (sa[i].len - sa[sa[i].link].len);
  }
  cout << sum << endl;
  return 0;
}