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;
struct SuffixAutomaton {
  vector<unordered_map<char, int> > edges;
  vector<int> link;
  vector<int> length;
  vector<bool> terminal;
  int last;
  SuffixAutomaton() {}
  SuffixAutomaton(string s) {
    edges.push_back({});
    link.push_back(-1);
    length.push_back(0);
    last = 0;
    for (int i = 0; i < s.size(); i++) {
      edges.push_back({});
      length.push_back(i + 1);
      link.push_back(0);
      int r = edges.size() - 1;
      int p = last;
      while (p >= 0 && edges[p][s[i]] == 0) {
        edges[p][s[i]] = r;
        p = link[p];
      }
      if (p != -1) {
        int q = edges[p][s[i]];
        if (length[p] + 1 == length[q]) {
          link[r] = q;
        } else {
          edges.push_back(edges[q]);
          length.push_back(length[p] + 1);
          link.push_back(link[q]);
          int qq = edges.size() - 1;
          link[q] = qq;
          link[r] = qq;
          while (p >= 0 && edges[p][s[i]] == q) {
            edges[p][s[i]] = qq;
            p = link[p];
          }
        }
      }
      last = r;
    }
    terminal.resize(edges.size());
    int p = last;
    while (p > 0) {
      terminal[p] = true;
      p = link[p];
    }
  }
} sa;
long long dp[200010];
long long sol(int node) {
  long long &ans = dp[node];
  if (ans != -1) return ans;
  ans = 0;
  if (sa.terminal[node]) ans++;
  for (pair<char, int> i : sa.edges[node])
    if (i.second != 0) ans += sol(i.second);
  return ans;
}
void printsa() {
  for (int i = 0; i < sa.edges.size(); i++) {
    for (pair<char, int> j : sa.edges[i]) {
      cout << i << " " << j.second << " " << j.first << endl;
    }
  }
  for (int i = 0; i < sa.edges.size(); i++) {
    if (sa.terminal[i]) cout << i << endl;
  }
  for (int i = 0; i < sa.edges.size(); i++) {
    cout << i << " suf-link " << sa.link[i] << endl;
  }
  cout << "last " << sa.last << endl;
}
long long solve() {
  long long ans = 0;
  for (int i = 1; i < sa.edges.size(); i++) {
    long long c = sol(i);
    long long c1 = sa.length[i] - sa.length[sa.link[i]];
    long long aux = c * (c + 1);
    aux = aux / 2;
    ans += aux * c1;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  sa = SuffixAutomaton(s);
  memset(dp, -1, sizeof dp);
  cout << solve() << endl;
  return 0;
}