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 SEED = chrono::steady_clock::now().time_since_epoch().count();
mt19937 rng(SEED);
const int MAX = 100000;
struct state {
  int len, link;
  map<char, int> next;
};
state st[2 * MAX];
int cnt[2 * MAX];
int sz, last;
void init() {
  sz = 1;
  last = 0;
  st[0].len = 0;
  st[0].link = -1;
  st[0].next.clear();
}
void extend(char c) {
  int cur = sz++;
  st[cur].len = st[last].len + 1;
  st[cur].next.clear();
  cnt[cur] = 1;
  int p;
  for (p = last; p != -1 && !st[p].next.count(c); p = st[p].link)
    st[p].next[c] = cur;
  if (p == -1)
    st[cur].link = 0;
  else {
    int q = st[p].next[c];
    if (st[p].len + 1 == st[q].len)
      st[cur].link = q;
    else {
      int clone = sz++;
      st[clone].len = st[p].len + 1;
      st[clone].next = st[q].next;
      st[clone].link = st[q].link;
      for (; p != -1 && st[p].next[c] == q; p = st[p].link)
        st[p].next[c] = clone;
      st[q].link = st[cur].link = clone;
    }
  }
  last = cur;
}
void test_case() {
  init();
  string str;
  cin >> str;
  int n = str.size();
  for (auto c : str) extend(c);
  vector<long long int> first(n + 1);
  for (int i = 1; i <= n; i++) first[i] = first[i - 1] + i;
  vector<int> order(sz);
  for (int i = 0; i < sz; i++) order[i] = i;
  sort(order.begin(), order.end(),
       [&](int i, int j) { return st[i].len > st[j].len; });
  for (auto u : order)
    if (u) cnt[st[u].link] += cnt[u];
  long long int res = 0;
  for (int u = 1; u < sz; u++)
    res += (st[u].len - st[st[u].link].len) * first[cnt[u]];
  cout << res << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    test_case();
  }
  return 0;
}