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 = 100005;
const int INF = 0x3f3f3f3f;
const int CHAR_SET = 30;
class suffix_automaton {
 private:
  vector<vector<int> > ch;
  vector<int> link, len, sz, cnt, rk;
  int n, last;
  int get_chr(int c) const { return c - 'a'; }

 public:
  suffix_automaton(int _n = N)
      : ch(_n, vector<int>(CHAR_SET)),
        link(_n),
        len(_n),
        sz(_n),
        cnt(_n),
        rk(_n),
        n(1),
        last(1) {}
  int extend(int c) {
    c = get_chr(c);
    int p = last, np = ++n;
    len[np] = len[p] + 1;
    for (; p && !ch[p][c]; p = link[p]) ch[p][c] = np;
    if (!p)
      link[np] = 1;
    else {
      int q = ch[p][c];
      if (len[q] == len[p] + 1)
        link[np] = q;
      else {
        int nq = ++n;
        ch[nq] = ch[q];
        link[nq] = link[q];
        len[nq] = len[p] + 1;
        link[q] = link[np] = nq;
        for (; p && ch[p][c] == q; p = link[p]) ch[p][c] = nq;
      }
    }
    sz[last = np]++;
    return len[last] - len[link[last]];
  }
  int size() const { return n; }
  const vector<int>& calc_size() {
    for (int i = 1; i <= n; i++) cnt[len[i]]++;
    for (int i = 1; i <= n; i++) cnt[i] += cnt[i - 1];
    for (int i = 1; i <= n; i++) rk[cnt[len[i]]--] = i;
    for (int i = n; i >= 1; i--) sz[link[rk[i]]] += sz[rk[i]];
    return sz;
  }
  long long solve() {
    long long ans = 0;
    for (int i = n; i > 1; i--)
      ans += (long long)(len[i] - len[link[i]]) * sz[i] * (sz[i] + 1) / 2;
    return ans;
  }
};
char s[N];
int main() {
  while (~scanf("%s", s)) {
    int n = strlen(s);
    suffix_automaton sa(n * 2 + 5);
    for (int i = 0; i < n; i++) sa.extend(s[i]);
    sa.calc_size();
    printf("%lld\n", sa.solve());
  }
  return 0;
}