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 n, last, cnt;
int bein[200005], leaf[200005];
struct SAM {
  int nex[26], link, len;
} A[200005];
void Insert(int c) {
  int p, cur = ++cnt;
  A[cur].len = A[last].len + 1;
  for (p = last; p != -1 && !A[p].nex[c]; p = A[p].link) A[p].nex[c] = cur;
  last = cur;
  leaf[cur] = 1;
  if (p == -1)
    A[cur].link = 0;
  else {
    int q = A[p].nex[c];
    if (A[q].len == A[p].len + 1)
      A[cur].link = q;
    else {
      int clone = ++cnt;
      A[clone] = A[q];
      A[clone].len = A[p].len + 1;
      for (; p != -1 && A[p].nex[c] == q; p = A[p].link) A[p].nex[c] = clone;
      A[cur].link = A[q].link = clone;
    }
  }
}
queue<int> q;
int main() {
  scanf("%s", S);
  n = strlen(S);
  last = cnt = 0;
  A[0].link = -1;
  for (int i = 0; i < n; i++) Insert(S[i] - 'a');
  for (int i = 1; i <= cnt; i++) bein[A[i].link]++;
  for (int i = 1; i <= cnt; i++)
    if (!bein[i]) q.push(i);
  long long ans = 0;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    if (!x) break;
    bein[A[x].link]--;
    if (!bein[A[x].link]) q.push(A[x].link);
    leaf[A[x].link] += leaf[x];
    ans +=
        (long long)leaf[x] * (leaf[x] + 1) / 2 * (A[x].len - A[A[x].link].len);
  }
  printf("%I64d\n", ans);
}