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 x[100005];
int n, a = 1, cnt = 1, la = 1;
int fa[200005], nex[200005][26], len[200005], v[200005], u[200005], sg[200005];
long long ans;
inline void extend(int c) {
  int p = la, np = la = ++cnt;
  len[np] = len[p] + 1;
  for (; p && !nex[p][c]; p = fa[p]) nex[p][c] = np;
  if (!p)
    fa[np] = 1;
  else {
    int q = nex[p][c], nq;
    if (len[q] == len[p] + 1)
      fa[np] = q;
    else {
      len[nq = ++cnt] = len[p] + 1;
      fa[nq] = fa[q];
      fa[q] = fa[np] = nq;
      memcpy(nex[nq], nex[q], sizeof nex[q]);
      for (; nex[p][c] == q; p = fa[p]) nex[p][c] = nq;
    }
  }
}
int main() {
  scanf("%s", x + 1);
  n = strlen(x + 1);
  for (int i = 1; i <= n; ++i) extend(u[i] = x[i] - 'a');
  for (int i = 1; i <= n; ++i) a = nex[a][u[i]], ++sg[a];
  for (int i = 1; i <= cnt; ++i) ++v[len[i]];
  for (int i = 1; i <= cnt; ++i) v[i] += v[i - 1];
  for (int i = 1; i <= cnt; ++i) u[v[len[i]]--] = i;
  for (int i = cnt; i >= 1; --i) sg[fa[u[i]]] += sg[u[i]];
  for (int i = cnt; i >= 1; --i)
    ans += (long long)(len[i] - len[fa[i]]) * sg[i] * (sg[i] + 1) / 2;
  printf("%I64d\n", ans);
  return 0;
}