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;
long long MOD = 1e9 + 7;
long long MAX = 1e17;
int a[21][100005];
long long n;
pair<pair<int, int>, int> p[100005];
long long lcp(int x, int y) {
  long long ans = 0;
  for (int stp = ceil(log2(n)); stp > -1; stp--) {
    if (a[stp][x] == a[stp][y]) {
      x += (1 << stp);
      y += (1 << stp);
      ans += (1 << stp);
    }
    if (x == n or y == n) break;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  ;
  string s;
  cin >> s;
  n = s.size();
  for (size_t i = 0; i < s.size(); i++) {
    a[0][i] = s[i] - 'a';
  }
  long long ans = (n * (n + 1)) / 2;
  int stp = 1, cnt = 1;
  for (; cnt < n; stp++, cnt *= 2) {
    for (int i = 0; i < n; i++) {
      p[i].first.first = a[stp - 1][i];
      p[i].first.second = (i + cnt < n) ? a[stp - 1][i + cnt] : -1;
      p[i].second = i;
    }
    sort(p, p + n);
    for (int i = 0; i < n; i++) {
      a[stp][p[i].second] =
          (i > 0 and p[i].first.first == p[i - 1].first.first and
           p[i].first.second == p[i - 1].first.second)
              ? a[stp][p[i - 1].second]
              : i;
    }
  }
  long long b[n - 1];
  for (int i = 0; i < n; i++) {
    p[i].first.first = a[stp - 1][i];
    p[i].first.second = -1;
    p[i].second = i;
  }
  sort(p, p + n);
  for (int i = 1; i < n; i++) {
    b[i - 1] = lcp(p[i - 1].second, p[i].second);
  }
  int k = 1, i = 0;
  long long f[n], w[n], h[n];
  f[0] = 0;
  h[0] = -1;
  w[0] = 0;
  while (i < n - 1) {
    long long l = 1;
    while (h[k - 1] >= b[i]) l += w[--k];
    w[k] = l;
    f[k] = f[k - 1] + l * (b[i]);
    h[k] = b[i];
    ans += f[k];
    k++;
    i++;
  }
  cout << ans << endl;
}