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;
int n;
char s[100005];
struct SaNode {
  int link, len, nrAp;
  int sons[26];
} sa[3 * 100005];
int last, cnt;
void addCharToSa(int c) {
  int crt = (++cnt);
  sa[crt].len = sa[last].len + 1;
  sa[crt].nrAp = 1;
  for (; last != -1 && sa[last].sons[c] == 0; last = sa[last].link)
    sa[last].sons[c] = crt;
  if (last == -1) {
    sa[crt].link = 0;
    last = crt;
    return;
  }
  int x = sa[last].sons[c];
  if (sa[last].len + 1 == sa[x].len)
    sa[crt].link = x;
  else {
    int clone = (++cnt);
    sa[clone].len = sa[last].len + 1;
    sa[clone].link = sa[x].link;
    sa[x].link = sa[crt].link = clone;
    sa[clone].nrAp = 0;
    for (int i = 0; i < 26; ++i) sa[clone].sons[i] = sa[x].sons[i];
    for (; last != -1 && sa[last].sons[c] == x; last = sa[last].link)
      sa[last].sons[c] = clone;
  }
  last = crt;
}
bool Comp(int a, int b) { return sa[a].len > sa[b].len; }
void buildSa() {
  sa[0].link = -1;
  for (int i = 0; i < n; ++i) addCharToSa(s[i] - 'a');
  n = cnt;
  int ord[3 * 100005];
  for (int i = 1; i <= n; ++i) ord[i] = i;
  sort(ord + 1, ord + n + 1, Comp);
  for (int i = 1; i <= n; ++i) {
    int x = ord[i];
    sa[sa[x].link].nrAp += sa[x].nrAp;
  }
}
int main() {
  scanf("%s", s);
  n = strlen(s);
  buildSa();
  long long sol = 0;
  for (int i = 1; i <= n; ++i) {
    int x = sa[i].nrAp;
    sol += 1LL * x * (x + 1) / 2 * (sa[i].len - sa[sa[i].link].len);
  }
  printf("%I64d\n", sol);
  return 0;
}