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 SaNode {
  int length, link, nrAp;
  int sons[26];
};
SaNode sa[2 * 100005];
int last = 0, nNodes = 0, ord[2 * 100005];
long long ans;
string s;
void addToSa(int c) {
  int node = (++nNodes);
  int p;
  sa[node].length = sa[last].length + 1;
  sa[node].nrAp = 1;
  for (p = last; p >= 0 && !sa[p].sons[c]; p = sa[p].link) sa[p].sons[c] = node;
  if (p < 0) {
    last = node;
    return;
  }
  int q = sa[p].sons[c];
  if (sa[p].length + 1 == sa[q].length)
    sa[node].link = q;
  else {
    int clone = (++nNodes);
    sa[clone].length = sa[p].length + 1;
    for (; p >= 0 && sa[p].sons[c] == q; p = sa[p].link) sa[p].sons[c] = clone;
    for (int i = 0; i < 26; ++i) sa[clone].sons[i] = sa[q].sons[i];
    sa[clone].link = sa[q].link;
    sa[q].link = sa[node].link = clone;
  }
  last = node;
}
long long f(int x) { return 1LL * x * (x + 1) / 2; }
void calcNrAp() {
  for (int i = 1; i <= nNodes; ++i) ord[i] = i;
  sort(ord + 1, ord + nNodes + 1,
       [&](int a, int b) { return sa[a].length > sa[b].length; });
  for (int i = 1; i <= nNodes; ++i) {
    int x = ord[i];
    sa[sa[x].link].nrAp += sa[x].nrAp;
    int numSubstrings = sa[x].length - sa[sa[x].link].length;
    ans += f(sa[x].nrAp) * numSubstrings;
  }
}
int main() {
  ios::sync_with_stdio(0);
  sa[0].link = -1;
  cin >> s;
  for (char c : s) addToSa(c - 'a');
  calcNrAp();
  cout << ans << "\n";
  return 0;
}