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 maxN = 200001;
int n;
string s;
int c[maxN];
int p[maxN];
int cn[maxN];
int where[maxN];
int who[maxN];
int lcp[maxN];
vector<int> g;
vector<int> x[maxN];
int k = 0;
int h[maxN];
int w[maxN];
long long f[maxN];
int main() {
  cin >> s;
  s += 'a' - 1;
  n = (int)s.size();
  vector<pair<char, int> > v;
  for (int i = 0; i < n; i++) {
    v.push_back(make_pair(s[i], i));
  }
  sort(v.begin(), v.end());
  int classes = 0;
  for (int i = 0; i < n; i++) {
    if (i == 0 || v[i].first != v[i - 1].first) {
      classes++;
    }
    c[v[i].second] = classes - 1;
    p[i] = v[i].second;
  }
  for (int h = 0; (1 << h) < n; h++) {
    k = (1 << h);
    vector<pair<pair<int, int>, int> > v1;
    for (int i = 0; i <= classes; i++) {
      x[i].clear();
    }
    for (int i = 0; i < n; i++) {
      x[c[(p[i] + k) % n]].push_back(p[i]);
    }
    vector<int> pn;
    for (int i = 0; i < classes; i++) {
      for (int j = 0; j < (int)x[i].size(); j++) {
        pn.push_back(x[i][j]);
      }
    }
    for (int i = 0; i < n; i++) {
      p[i] = pn[i];
    }
    for (int i = 0; i <= classes; i++) {
      x[i].clear();
    }
    for (int i = 0; i < n; i++) {
      x[c[p[i]]].push_back(p[i]);
    }
    pn.clear();
    for (int i = 0; i < classes; i++) {
      for (int j = 0; j < (int)x[i].size(); j++) {
        pn.push_back(x[i][j]);
      }
    }
    for (int i = 0; i < n; i++) {
      p[i] = pn[i];
    }
    vector<int> cn;
    classes = 0;
    cn.resize(n);
    for (int i = 0; i < n; i++) {
      if (i == 0 || (c[p[i]] != c[p[i - 1]] ||
                     (c[(p[i] + k) % n] != c[(p[i - 1] + k) % n]))) {
        classes++;
      }
      cn[p[i]] = classes - 1;
    }
    for (int i = 0; i < n; i++) {
      c[i] = cn[i];
    }
  }
  for (int i = 0; i < n; i++) {
    where[p[i]] = i;
    who[i] = p[i];
  }
  int x = 0;
  for (int i = 0; i < n; i++) {
    x = max(0, x - 1);
    if (where[i] != n - 1) {
      int j = who[where[i] + 1];
      for (int g = i + x; g < n; g++) {
        if ((g - i + 1 > n - j)) {
          break;
        }
        if (s[g] != s[j + (g - i)]) {
          break;
        }
        x++;
      }
      lcp[where[i]] = x;
    }
  }
  long long ans = 1LL * (n - 1) * n / 2LL;
  int k = n - 2;
  n = 0;
  h[n] = -1;
  w[n] = 0;
  f[n] = 0;
  n++;
  for (int i = 0; i < k; i++) {
    int l = 1;
    while (h[n - 1] >= lcp[i + 1]) {
      l += w[--n];
    }
    w[n] = l;
    h[n] = lcp[i + 1];
    f[n] = f[n - 1] + 1LL * l * lcp[i + 1];
    n++;
    ans += f[n - 1];
  }
  cout << ans << endl;
  return 0;
}