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;
char s[200005];
int step[200005], son[200005][26], size[200005], f[200005];
int i, j, k, n, cnt, last = 1, tot = 1;
vector<int> a[200005];
long long tmp, ans;
inline void Insert(int ch) {
  int np = ++tot, p, q;
  step[np] = step[last] + 1;
  for (p = last; p && !son[p][ch]; p = f[p]) son[p][ch] = np;
  last = np;
  if (!p) {
    f[np] = 1;
    return;
  }
  if (step[q = son[p][ch]] == step[p] + 1) {
    f[np] = q;
    return;
  }
  int nq = ++tot;
  step[nq] = step[p] + 1;
  memcpy(son[nq], son[q], sizeof(son[q]));
  f[nq] = f[q];
  f[np] = f[q] = nq;
  for (; p && son[p][ch] == q; p = f[p]) son[p][ch] = nq;
}
void DFS(int x) {
  for (int i = 0; i < a[x].size(); i++) DFS(a[x][i]), size[x] += size[a[x][i]];
  tmp = 1ll * size[x] * (size[x] + 1) / 2ll;
  ans += 1ll * (step[x] - step[f[x]]) * tmp;
}
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (i = 1; i <= n; i++) Insert(s[i] - 'a'), size[last]++;
  for (i = 2; i <= tot; i++) a[f[i]].push_back(i);
  DFS(1);
  printf("%I64d\n", ans);
}