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 long long N = 1e5 + 5, inf = 1e9 + 5;
long long sa[N], lcp[N], gap, tmp[N], n, pos[N], r[N], l[N];
long long ans = 0;
string s;
bool sufcmp(long long x, long long y) {
  if (pos[x] != pos[y]) return pos[x] < pos[y];
  if (max(x, y) + gap < n) return pos[x + gap] < pos[y + gap];
  return x > y;
}
void buildSA() {
  for (long long i = 0; i < n; i++) {
    sa[i] = i;
    pos[i] = s[i];
  }
  for (gap = 1;; gap *= 2) {
    sort(sa, sa + n, sufcmp);
    for (long long i = 0; i < n - 1; i++)
      tmp[i + 1] = tmp[i] + sufcmp(sa[i], sa[i + 1]);
    for (long long i = 0; i < n; i++) pos[sa[i]] = tmp[i];
    if (tmp[n - 1] == n - 1) break;
  }
}
void buildLCP() {
  for (long long i = 0, k = 0; i < n; i++)
    if (pos[i] != n - 1) {
      for (long long j = sa[pos[i] + 1]; s[i + k] == s[j + k];) k++;
      lcp[pos[i]] = k;
      if (k) k--;
    }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  n = s.size();
  buildSA();
  buildLCP();
  lcp[n - 1] = n;
  l[0] = -1;
  for (long long i = 1; i < n - 1; i++) {
    l[i] = i - 1;
    while (l[i] >= 0 && lcp[l[i]] >= lcp[i]) l[i] = l[l[i]];
  }
  stack<long long> st;
  for (long long i = n - 2; i >= 0; i--) {
    while (st.size() && lcp[st.top()] > lcp[i]) st.pop();
    if (!st.size())
      r[i] = n - 1;
    else
      r[i] = st.top();
    st.push(i);
  }
  for (long long i = 0; i < n - 1; i++) {
    ans += ((i - l[i]) * (r[i] - i)) * lcp[i];
  }
  cout << ans + n * (n + 1) / 2 << '\n';
}