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 MAXN = 100000;
struct TNode {
  int suf;
  int len;
  int sufst;
  TNode() : suf(-1), len(0), sufst(0) {}
};
TNode st[3 * MAXN];
map<char, int> linx[3 * MAXN];
int cnt[3 * MAXN];
int stsize = 0;
int newNode() {
  st[stsize] = TNode();
  linx[stsize].clear();
  return stsize++;
}
int append(int end, char c) {
  int newEnd = newNode();
  st[newEnd] = TNode();
  cnt[newEnd] = 1;
  linx[end][c] = newEnd;
  st[newEnd].len = st[end].len + 1;
  st[newEnd].sufst = 0;
  int cur = st[end].suf;
  while (cur >= 0 && !linx[cur].count(c)) {
    linx[cur][c] = newEnd;
    cur = st[cur].suf;
  }
  if (cur < 0) {
    st[newEnd].suf = 0;
    return newEnd;
  }
  int q = linx[cur][c];
  if (st[q].len - st[cur].len == 1) {
    st[newEnd].suf = q;
    return newEnd;
  }
  int clone = newNode();
  st[clone] = st[q];
  linx[clone] = linx[q];
  st[clone].len = st[cur].len + 1;
  st[clone].sufst = st[newEnd].len - st[clone].len;
  st[newEnd].suf = clone;
  st[q].suf = clone;
  while (cur >= 0 && linx[cur].count(c) && linx[cur][c] == q) {
    linx[cur][c] = clone;
    cur = st[cur].suf;
  }
  return newEnd;
}
bool isSubstr(int start, char *s) {
  for (int i = 0; s[i]; ++i) {
    if (!linx[start].count(s[i])) {
      return false;
    }
    start = linx[start][s[i]];
  }
  return true;
}
char s[200000];
int main() {
  scanf("%s", s);
  int end = newNode();
  for (int i = 0; s[i]; ++i) {
    end = append(end, s[i]);
  }
  vector<pair<int, int> > ord;
  for (int i = 0; i < stsize; ++i) {
    ord.push_back(make_pair(st[i].len, i));
  }
  sort(ord.rbegin(), ord.rend());
  for (int i = 0; i < stsize; ++i) {
    int q = ord[i].second;
    cnt[st[q].suf] += cnt[q];
  }
  long long ans = 0;
  for (int i = 1; i < stsize; ++i) {
    int q = st[i].len - st[st[i].suf].len;
    ans += 1LL * q * cnt[i] * (cnt[i] + 1) / 2;
  }
  cout << ans << '\n';
  return 0;
}