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 maxx = 4e5 + 1314;
int tr[maxx][27];
int idx(char a) { return a - 'a' + 1; }
int now = 1, cnt = 1, par[maxx], mx[maxx], dp[maxx];
int lin[maxx], len;
struct edge {
  int y, next;
} e[maxx];
void init(int x, int y) {
  e[++len].y = y;
  e[len].next = lin[x];
  lin[x] = len;
}
void extend(int x) {
  int np = ++cnt, p = now;
  mx[np] = mx[p] + 1;
  now = np;
  while (!tr[p][x] && p) tr[p][x] = np, p = par[p];
  if (!p)
    par[np] = 1;
  else {
    int q = tr[p][x];
    if (mx[q] == mx[p] + 1)
      par[np] = q;
    else {
      int nq = ++cnt;
      mx[nq] = mx[p] + 1;
      par[nq] = par[q];
      par[q] = par[np] = nq;
      memcpy(tr[nq], tr[q], sizeof(tr[q]));
      while (tr[p][x] == q && p) tr[p][x] = nq, p = par[p];
    }
  }
  dp[np]++;
}
void dfs(int x) {
  for (int i = lin[x]; i; i = e[i].next) {
    int y = e[i].y;
    dfs(y);
    dp[x] += dp[y];
  }
}
long long ans = 0;
void slove() {
  for (int i = 2; i <= cnt; i++) init(par[i], i);
  dfs(1);
  for (int i = 2; i <= cnt; i++) {
    ans += 1ll * (mx[i] - mx[par[i]]) * (1 + dp[i]) * dp[i] / 2;
  }
  cout << ans << endl;
}
char s[maxx];
int main() {
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  for (int i = 1; i <= n; i++) extend(idx(s[i]));
  slove();
  return 0;
}