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 MAXN = 1e5 + 10;
const int LG = 20;
const int MOD = 1e9 + 7;
string s;
pair<pair<int, int>, int> arr[MAXN];
int sufrank[LG][MAXN], n;
int revrank[MAXN], lcp[MAXN];
int L[MAXN], R[MAXN];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  n = s.size();
  revrank[n] = -1;
  for (int i = 0; i < n; i++) sufrank[0][i] = s[i] - 'a' + 256;
  for (int l = 1; l < LG; l++) {
    for (int i = 0, j = 1 << (l - 1); i < n; i++, j++)
      arr[i] = {{sufrank[l - 1][i], j < n ? sufrank[l - 1][j] : -256}, i};
    sort(arr, arr + n);
    for (int i = 1; i < n; i++)
      sufrank[l][arr[i].second] = (arr[i].first == arr[i - 1].first)
                                      ? sufrank[l][arr[i - 1].second]
                                      : sufrank[l][arr[i - 1].second] + 1;
  }
  for (int i = 0; i < n; i++) revrank[sufrank[LG - 1][i]] = i;
  for (int i = 0, E = 0; i < n; i++) {
    int A = revrank[sufrank[LG - 1][i] + 1];
    if (A == -1) {
      E = 0;
      continue;
    }
    while ((A + E) < n && (i + E) < n && s[A + E] == (s[i + E])) E++;
    lcp[sufrank[LG - 1][i]] = E;
    if (E) E--;
  }
  vector<int> vec;
  for (int i = 0; i < (n - 1); i++) {
    while (vec.size() && lcp[vec.back()] > lcp[i]) vec.pop_back();
    L[i] = vec.size() ? vec.back() : -1;
    vec.push_back(i);
  }
  vec.clear();
  for (int i = n - 2; ~i; i--) {
    while (vec.size() && lcp[vec.back()] >= lcp[i]) vec.pop_back();
    R[i] = vec.size() ? vec.back() : n - 1;
    vec.push_back(i);
  }
  long long answer = 0;
  for (int i = 0; i < (n - 1); i++) {
    answer += 1ll * (lcp[i] + 1) * (i - L[i]) * (R[i] - i);
  }
  cout << answer + n << endl;
  return 0;
}