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;
class SuffixDataStructures {
 public:
  int n;
  string s;
  vector<int> sa;
  vector<int> rank;
  vector<int> lcp;
  SuffixDataStructures(const string& s) : s(s) {
    n = s.size();
    buildsa();
    buildlcp();
  }
  void buildsa() {
    sa.resize(n);
    rank.resize(n);
    for (int i = 0; i < n; i++) {
      sa[i] = i;
    }
    sort(sa.begin(), sa.end(), [&](int x, int y) { return s[x] < s[y]; });
    vector<int> oldrank(n);
    for (int len = 1;; len *= 2) {
      int cnt = 0;
      swap(rank, oldrank);
      for (int i = 0; i < n; i++) {
        int x = sa[i];
        bool eq;
        if (i == 0) {
          eq = false;
        } else {
          int w = sa[i - 1];
          if (len == 1) {
            eq = s[w] == s[x];
          } else {
            eq = oldrank[w] == oldrank[x] &&
                 oldrank[w + len / 2] == oldrank[x + len / 2];
          }
        }
        if (!eq) {
          cnt++;
        }
        rank[x] = cnt - 1;
      }
      if (cnt == n) {
        break;
      }
      sort(sa.begin(), sa.end(), [&](int x, int y) {
        return rank[x] != rank[y] ? rank[x] < rank[y]
                                  : rank[x + len] < rank[y + len];
      });
    }
  }
  void buildlcp() {
    lcp.resize(n);
    for (int i = 0, len = 0; i < n - 1; i++) {
      int h = sa[rank[i] - 1];
      while (s[h + len] == s[i + len]) {
        len++;
      }
      lcp[rank[i]] = len;
      if (len > 0) {
        len--;
      }
    }
  }
};
int main() {
  ios::sync_with_stdio(false);
  string s;
  cin >> s;
  s += '\0';
  SuffixDataStructures S(s);
  vector<pair<int, int>> a;
  long long ans = 0;
  for (int i = 0; i < S.n; i++) {
    while (!a.empty() && S.lcp[i] <= a.back().first) {
      a.pop_back();
    }
    a.emplace_back(S.lcp[i], i - 1);
    int z = i == S.n - 1 ? 0 : S.lcp[i + 1];
    for (auto it = a.rbegin(); it != a.rend(); ++it) {
      int w = max(it->first, z);
      int x;
      if (it == a.rbegin()) {
        x = S.n - S.sa[i] - w;
      } else {
        x = (it - 1)->first - w;
      }
      int y = i - it->second;
      ans += (long long)x * y * (y + 1) / 2;
      if (w <= z) {
        break;
      }
    }
  }
  ans -= S.n;
  printf("%lld\n", ans);
  return 0;
}