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 MaxN = 100010;
struct Node {
  int len;
  long long cnt1, cnt2;
  Node *nxt[26], *fa;
} pool[MaxN * 2], *TAIL = pool + 1, *tail = pool, *op = pool, *q[MaxN * 2];
int n, d[MaxN * 2];
char st[MaxN];
void extend(int c, int len) {
  Node *p = tail, *np = tail = TAIL++;
  np->len = len;
  for (; p && !p->nxt[c]; p = p->fa) p->nxt[c] = np;
  if (!p)
    np->fa = op;
  else {
    if (p->nxt[c]->len == p->len + 1)
      np->fa = p->nxt[c];
    else {
      Node *q = p->nxt[c], *r = TAIL++;
      *r = *q;
      np->fa = q->fa = r;
      r->len = p->len + 1;
      for (; p && p->nxt[c] == q; p = p->fa) p->nxt[c] = r;
    }
  }
}
int main() {
  int l, r;
  long long ans = 0;
  scanf("%s", st + 1);
  n = strlen(st + 1);
  for (int i = 1; i <= n; i++) extend(st[i] - 'a', i);
  for (Node *p = pool; p < TAIL; p++)
    for (int i = 0; i < 26; i++)
      if (p->nxt[i]) d[((p->nxt[i]) - pool)]++;
  op->cnt1 = 1;
  for (q[l = r = 0] = op; l <= r; l++) {
    Node *p = q[l];
    for (int i = 0; i < 26; i++)
      if (p->nxt[i])
        p->nxt[i]->cnt1 += p->cnt1,
            (!(--d[((p->nxt[i]) - pool)]) ? q[++r] = p->nxt[i] : 0);
  }
  for (Node *p = tail; p; p = p->fa) p->cnt2 = 1;
  for (; r; r--) {
    Node *p = q[r];
    for (int i = 0; i < 26; i++)
      if (p->nxt[i]) p->cnt2 += p->nxt[i]->cnt2;
    ans += p->cnt1 * p->cnt2 * (p->cnt2 + 1) / 2;
  }
  cout << ans << endl;
  return 0;
}