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;
struct node {
  node *par, *go[26];
  int num;
  int val;
} * root, *tail, que[200010], *top[200010];
int tot;
char str[100010];
void add(int c, int l) {
  node *p = tail, *np = &que[tot++];
  np->val = l;
  while (p && p->go[c] == NULL) p->go[c] = np, p = p->par;
  if (p == NULL)
    np->par = root;
  else {
    node *q = p->go[c];
    if (p->val + 1 == q->val)
      np->par = q;
    else {
      node *nq = &que[tot++];
      *nq = *q;
      nq->val = p->val + 1;
      np->par = q->par = nq;
      while (p && p->go[c] == q) p->go[c] = nq, p = p->par;
    }
  }
  tail = np;
}
int c[200010], len;
void init() {
  memset(que, 0, sizeof(que));
  tot = 0;
  len = 1;
  root = tail = &que[tot++];
}
void solve() {
  memset(c, 0, sizeof(c));
  int i;
  for (i = 0; i < tot; i++) c[que[i].val]++;
  for (i = 1; i < len; i++) c[i] += c[i - 1];
  for (i = 0; i < tot; i++) top[--c[que[i].val]] = &que[i];
  for (node *p = root;; p = p->go[str[p->val] - 'a']) {
    p->num = 1;
    if (p->val == len - 1) break;
  }
  for (i = tot - 1; i >= 0; i--) {
    node *p = top[i];
    if (p->par) {
      node *q = p->par;
      q->num += p->num;
    }
  }
  long long ans = 0;
  for (i = 1; i < tot; i++) {
    node *p = top[i];
    int tmp = p->val - p->par->val, sum = p->num;
    ans += (long long)tmp * ((long long)sum * (sum + 1) / 2);
  }
  printf("%I64d\n", ans);
}
int main() {
  scanf("%s", str);
  int l = strlen(str), i;
  init();
  for (i = 0; i < l; i++) {
    add(str[i] - 'a', len++);
  }
  solve();
  return 0;
}