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;
int const N = 1e5 + 10;
int r[N << 1], id[N << 1];
bool v[N << 1];
char s[N];
vector<pair<int, int>> o;
struct SAM {
  int sz, last;
  struct state {
    int len, link, num;
    int nxt[26];
  } st[N << 1];
  void init() {
    sz = last = 0;
    st[0].len = 0, st[0].link = -1, st[0].num = 0;
    ++sz;
    for (int i = 0; i < N * 2; ++i) memset(st[i].nxt, 0, sizeof(int) * 26);
  }
  void extend(int c) {
    int cur = sz++;
    st[cur].num = 1;
    st[cur].len = st[last].len + 1;
    int p;
    for (p = last; p != -1 && !st[p].nxt[c]; p = st[p].link) st[p].nxt[c] = cur;
    if (p == -1) {
      st[cur].link = 0;
    } else {
      int q = st[p].nxt[c];
      if (st[p].len + 1 == st[q].len) {
        st[cur].link = q;
      } else {
        int clone = sz++;
        st[clone].num = 0;
        st[clone].len = st[p].len + 1;
        st[clone].link = st[q].link;
        memcpy(st[clone].nxt, st[q].nxt, sizeof(int) * 26);
        for (; p != -1 && st[p].nxt[c] == q; p = st[p].link)
          st[p].nxt[c] = clone;
        st[q].link = st[cur].link = clone;
      }
    }
    last = cur;
  }
  void calc() {
    for (int i = 0; i < sz; ++i) ++r[st[i].len];
    partial_sum(r, r + sz, r);
    for (int i = 0; i < sz; ++i) id[--r[st[i].len]] = i;
    for (int i = sz - 1; i >= 0; --i)
      if (st[id[i]].link != -1) {
        st[st[id[i]].link].num += st[id[i]].num;
      }
  }
  long long gao() {
    long long tot = 0;
    int p = 0;
    for (int i = 0; s[i]; ++i) {
      int c = s[i] - 'a';
      p = st[p].nxt[c];
      for (int j = p; j > 0; j = st[j].link) {
        if (v[j]) break;
        v[j] = true;
        int k = st[j].link;
        long long cnt = st[j].len - st[k].len;
        tot += cnt * (st[j].num + 1ll) * st[j].num / 2;
      }
    }
    return tot;
  }
} sam;
int main() {
  scanf("%s", s);
  sam.init();
  for (int i = 0; s[i]; ++i) sam.extend(s[i] - 'a');
  sam.calc();
  printf("%I64d\n", sam.gao());
  return 0;
}