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 Max = 1e5 + 10;
const int LOG = 20;
string s;
int n;
int rnk[LOG][Max];
pair<pair<int, int>, int> sorter[Max];
int L[Max];
int R[Max];
int pos[Max];
int lcp[Max];
void build() {
  for (int i = 0; i < n; i++) rnk[0][i] = s[i];
  for (int lg = 1, sz = 1; lg < LOG; lg++, sz <<= 1) {
    for (int i = 0; i < n; i++)
      sorter[i] = make_pair(
          make_pair(rnk[lg - 1][i], i + sz < n ? rnk[lg - 1][i + sz] : -1), i);
    sort(sorter, sorter + n);
    for (int i = 0, j = 0; i < n; i++)
      rnk[lg][sorter[i].second] = j =
          j + (i > 0 && sorter[i].first != sorter[i - 1].first);
  }
  for (int i = 0; i < n; i++) pos[rnk[LOG - 1][i]] = i;
  for (int lg = LOG; lg--;)
    for (int i = 0; i < n; i++)
      if (max(pos[i], pos[i + 1]) + lcp[i] < n &&
          rnk[lg][pos[i] + lcp[i]] == rnk[lg][pos[i + 1] + lcp[i]])
        lcp[i] |= (1 << lg);
}
int main() {
  cin >> s;
  n = s.size();
  build();
  vector<pair<int, int> > vec;
  for (int i = 0; i < n - 1; i++) {
    while (vec.size() && vec.back().first >= lcp[i]) vec.pop_back();
    if (vec.size())
      L[i] = vec.back().second + 1;
    else
      L[i] = 0;
    vec.push_back(make_pair(lcp[i], i));
  }
  vec.clear();
  for (int i = n - 1; i--;) {
    while (vec.size() && vec.back().first > lcp[i]) vec.pop_back();
    if (vec.size())
      R[i] = vec.back().second;
    else
      R[i] = n - 1;
    vec.push_back(make_pair(lcp[i], i));
  }
  long long ANS = 0;
  for (int i = 0; i < n - 1; i++)
    ANS += 1ll * lcp[i] * (i - L[i] + 1) * (R[i] - i);
  ANS += 1ll * n * (n + 1) / 2;
  cout << ANS << '\n';
}