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>
struct node {
  int len, link, s;
  std::map<char, int> urm;
} sa[2 * 100000 + 1];
int last, sz;
long long ans;
std::vector<int> g[2 * 100000 + 1];
inline void init() {
  last = 0;
  sa[0].len = 0;
  sa[0].link = -1;
  sz = 1;
}
inline void baga(char x) {
  int cur = sz++;
  sa[cur].len = sa[last].len + 1;
  int p = last;
  while (p != -1 && sa[p].urm[x] == 0) {
    sa[p].urm[x] = cur;
    p = sa[p].link;
  }
  if (p == -1)
    sa[cur].link = 0;
  else {
    int q = sa[p].urm[x];
    if (sa[q].len == sa[p].len + 1)
      sa[cur].link = q;
    else {
      int clone = sz++;
      sa[clone].len = sa[p].len + 1;
      sa[clone].link = sa[q].link;
      sa[clone].urm = sa[q].urm;
      while (p != -1 && sa[p].urm[x] == q) {
        sa[p].urm[x] = clone;
        p = sa[p].link;
      }
      sa[q].link = sa[cur].link = clone;
    }
  }
  last = cur;
  sa[cur].s = 1;
}
void dfs(int x) {
  for (auto &y : g[x]) {
    dfs(y);
    sa[x].s += sa[y].s;
  }
}
int main() {
  init();
  char ch = fgetc(stdin);
  while (ch != '\n') {
    baga(ch);
    ch = fgetc(stdin);
  }
  for (int i = 1; i < sz; i++) g[sa[i].link].push_back(i);
  dfs(0);
  long long ans = 0;
  for (int i = 1; i < sz; i++)
    ans += 1LL * (sa[i].len - sa[sa[i].link].len) *
           (1LL * sa[i].s * (sa[i].s + 1) / 2);
  fprintf(stdout, "%I64d\n", ans);
  fclose(stdin);
  fclose(stdout);
  return 0;
}