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 = 1e6 + 10;
const int INF = 1e9 + 9;
const int B = 1e9 + 7;
const int alph = 26;
struct node {
  int len, cnt;
  int to[alph], link;
};
struct SufAuto {
  int sz, last;
  node atm[N];
  void insert(int ch) {
    int new_node = ++sz;
    atm[new_node].len = atm[last].len + 1;
    atm[new_node].cnt = 1;
    int ptr = last;
    for (; ptr && atm[ptr].to[ch] == 0; ptr = atm[ptr].link)
      atm[ptr].to[ch] = new_node;
    if (ptr == 0)
      atm[new_node].link = 1;
    else {
      int var = atm[ptr].to[ch];
      if (atm[var].len == atm[ptr].len + 1)
        atm[new_node].link = var;
      else {
        int clone = ++sz;
        atm[clone] = atm[var];
        atm[clone].cnt = 0;
        atm[clone].len = atm[ptr].len + 1;
        atm[var].link = atm[new_node].link = clone;
        for (; ptr && atm[ptr].to[ch] == var; ptr = atm[ptr].link)
          atm[ptr].to[ch] = clone;
      }
    }
    last = new_node;
  }
  vector<int> g[N];
  void dfs(int first) {
    for (auto to : g[first]) {
      dfs(to);
      atm[first].cnt += atm[to].cnt;
    }
  }
  void build_tree() {
    for (int i = int(1); i <= int(sz); ++i) g[atm[i].link].push_back(i);
    dfs(0);
  }
  void init(const string &s) {
    sz = last = 1;
    atm[last].link = 0;
    for (auto c : s) insert(c - 'a');
    build_tree();
  }
} sa;
string s;
long long c2(long long first) { return first * (first + 1) / 2; }
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cin >> s;
  sa.init(s);
  long long ans = 0;
  for (int i = int(1); i <= int(sa.sz); ++i) {
    int p = sa.atm[i].link;
    ans += c2(sa.atm[i].cnt) * (sa.atm[i].len - sa.atm[p].len);
  }
  cout << ans << "\n";
  return 0;
}