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 N = 1e5 + 7;
struct node {
  int son[26], pre, len;
} a[N << 1];
int tot, tail = 0, n;
char s[N];
long long ans = 0;
int sz[N << 1];
void ins(int k) {
  int p, q, np = ++tot, nq;
  a[np].len = a[tail].len + 1;
  for (p = tail; p && !a[p].son[k]; p = a[p].pre) a[p].son[k] = np;
  sz[np] = 1;
  if (!a[p].son[k]) {
    a[p].son[k] = np;
    a[np].pre = p;
  } else {
    q = a[p].son[k];
    if (a[q].len == a[p].len + 1)
      a[np].pre = q;
    else {
      a[nq = ++tot] = a[q];
      a[nq].len = a[p].len + 1;
      a[q].pre = a[np].pre = nq;
      for (; a[p].son[k] == q; p = a[p].pre) a[p].son[k] = nq;
    }
  }
  tail = np;
}
vector<int> g[N << 1];
void dfs(int u) {
  for (int v : g[u]) {
    dfs(v);
    sz[u] += sz[v];
    ans += 1LL * sz[v] * (sz[v] + 1) / 2 * (a[v].len - a[u].len);
  }
}
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (int i = int(n); i >= int(1); --i) ins(s[i] - 'a');
  for (int i = int(1); i <= int(tot); ++i) {
    int last = a[i].pre;
    g[last].push_back(i);
  }
  dfs(0);
  cout << ans << endl;
}