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;
string s;
int n, aa, a[101010], b[101010], c[101010], d[101010];
vector<pair<int, int> > v;
long long ans, sum;
bool cmp(int x, int y) {
  if (c[x] != c[y]) {
    return c[x] < c[y];
  }
  x += aa;
  y += aa;
  if (x < n && y < n) {
    return c[x] < c[y];
  } else {
    return x > y;
  }
}
int main() {
  cin >> s;
  n = s.size();
  for (int i = 0; i < n; i++) {
    b[i] = i;
    c[i] = s[i];
  }
  for (aa = 1;; aa *= 2) {
    sort(b, b + n, cmp);
    for (int i = 0; i < n - 1; i++) {
      d[i + 1] = d[i] + cmp(b[i], b[i + 1]);
    }
    for (int i = 0; i < n; i++) {
      c[b[i]] = d[i];
    }
    if (d[n - 1] == n - 1) {
      break;
    }
  }
  int nw = 0;
  for (int i = 0; i < n; i++) {
    if (c[i] != n - 1) {
      for (int j = b[c[i] + 1]; s[i + nw] == s[j + nw];) {
        nw++;
      }
      a[c[i]] = nw;
      nw = max(nw - 1, 0);
    }
  }
  for (int i = 0; i < n - 1; i++) {
    int t = 1;
    while (v.size() && v.back().first >= a[i]) {
      sum -= v.back().first * v.back().second;
      t += v.back().second;
      v.pop_back();
    }
    sum += a[i] * t;
    ans += sum;
    v.push_back(make_pair(a[i], t));
  }
  ans += 1ll * n * (n + 1) / 2;
  cout << ans << endl;
  return 0;
}