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 N = 200005;
int root, last, nodecnt;
int u[N][26], val[N], par[N];
inline int newnode(int _val) {
  ++nodecnt;
  val[nodecnt] = _val, par[nodecnt] = 0;
  return nodecnt;
}
void extend(int c) {
  int p = last, np = newnode(val[p] + 1);
  while (p && u[p][c] == 0) u[p][c] = np, p = par[p];
  if (p == 0)
    par[np] = root;
  else {
    int q = u[p][c];
    if (val[p] + 1 == val[q]) {
      par[np] = q;
    } else {
      int nq = newnode(val[p] + 1);
      memcpy(u[nq], u[q], sizeof(u[q]));
      par[nq] = par[q];
      par[q] = par[np] = nq;
      while (p && u[p][c] == q) u[p][c] = nq, p = par[p];
    }
  }
  last = np;
}
char str[N];
int cnt[N], a[N];
long long f[N];
void work() {
  nodecnt = 0;
  root = last = newnode(0);
  scanf("%s", str);
  for (int i = 0; str[i]; ++i) extend(str[i] - 'a');
  for (int i = 1; i <= nodecnt; ++i) ++cnt[val[i]];
  for (int i = 1; i <= nodecnt; ++i) cnt[i] += cnt[i - 1];
  for (int i = nodecnt; i > 0; --i) a[cnt[val[i]]--] = i;
  while (last) f[last] = 1, last = par[last];
  long long res = 0;
  for (int i = nodecnt; i > 0; --i) {
    int x = a[i];
    for (int j = 0; j < 26; ++j) {
      int y = u[x][j];
      f[x] += f[y];
    }
    res += f[x] * (f[x] + 1) / 2 * (val[x] - val[par[x]]);
  }
  cout << res << endl;
}
int main() {
  work();
  return 0;
}