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 = 100 * 1000 + 85 - 69;
int sa[maxn], pos[maxn], tmp[maxn], lcp[maxn], last[maxn], nexT[maxn],
    next2[maxn], gap;
long long int ans;
string s;
stack<int> stc;
bool sufCmp(int i, int j) {
  if (pos[i] != pos[j]) return pos[i] < pos[j];
  i += gap, j += gap;
  return (i < ((int)(s).size()) and j < ((int)(s).size())) ? pos[i] < pos[j]
                                                           : i > j;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> s;
  for (int i = 0; i < ((int)(s).size()); i++) sa[i] = i, pos[i] = s[i];
  for (gap = 1;; gap <<= 1) {
    sort(sa, sa + ((int)(s).size()), sufCmp);
    for (int i = 0; i + 1 < ((int)(s).size()); i++)
      tmp[i + 1] = tmp[i] + sufCmp(sa[i], sa[i + 1]);
    for (int i = 0; i < ((int)(s).size()); i++) pos[sa[i]] = tmp[i];
    if (tmp[((int)(s).size()) - 1] == ((int)(s).size()) - 1) break;
  }
  int match = 0;
  for (int i = 0; i < ((int)(s).size()); i++) {
    if (match) match--;
    if (!pos[i]) continue;
    int cur = i, pre = sa[pos[i] - 1];
    while (max(pre, cur) + match < ((int)(s).size()) and
           s[pre + match] == s[cur + match])
      match++;
    lcp[pos[i]] = match;
  }
  for (int i = 0; i < ((int)(s).size()); i++) {
    while (((int)(stc).size()) and lcp[stc.top()] >= lcp[i]) stc.pop();
    last[i] = (((int)(stc).size()) ? stc.top() : 0);
    stc.push(i);
  }
  while (((int)(stc).size())) stc.pop();
  for (int i = ((int)(s).size()) - 1; i >= 0; i--) {
    while (((int)(stc).size()) and lcp[stc.top()] >= lcp[i]) stc.pop();
    nexT[i] = (((int)(stc).size()) ? stc.top() : ((int)(s).size()));
    stc.push(i);
  }
  while (((int)(stc).size())) stc.pop();
  for (int i = ((int)(s).size()) - 1; i >= 0; i--) {
    while (((int)(stc).size()) and lcp[stc.top()] > lcp[i]) stc.pop();
    next2[i] = (((int)(stc).size()) ? stc.top() : ((int)(s).size()));
    stc.push(i);
  }
  for (int i = 0; i < ((int)(s).size()); i++)
    if (lcp[i] and lcp[next2[i]] != lcp[i]) {
      long long int parh = max(lcp[last[i]], lcp[nexT[i]]);
      long long int dcnt = nexT[i] - last[i];
      ans += dcnt * (dcnt + 1) / 2 * (lcp[i] - parh);
    }
  for (int i = 0; i < ((int)(s).size()); i++) {
    int an = max(lcp[i], lcp[i + 1]);
    if (an < ((int)(s).size()) - sa[i]) ans += ((int)(s).size()) - sa[i] - an;
  }
  cout << ans << endl;
  return 0;
}