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 = 1e6 + 100;
const int MAXL = 25;
int rk[MAXL][MAXN], p[MAXN];
pair<pair<int, int>, int> tmp[MAXN];
int l[MAXN];
vector<int> g;
bool cmp(int a, int b) { return rk[MAXL - 1][a] < rk[MAXL - 1][b]; }
int lcp(int a, int b, int n) {
  int res = 0;
  if (a < b) swap(a, b);
  for (int i = MAXL - 1; i >= 0; i--) {
    if (a + (1 << i) > n) continue;
    if (rk[i][a] == rk[i][b]) {
      a += (1 << i);
      b += (1 << i);
      res += (1 << i);
    }
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  int n = s.size();
  for (int i = 0; i < n; i++) rk[0][i] = s[i] - 'a', p[i] = i;
  for (int bt = 1; bt < MAXL; bt++) {
    for (int i = 0; i < n; i++) {
      tmp[i] = {{rk[bt - 1][i], -1}, i};
      if ((i + (1 << (bt - 1))) < n)
        tmp[i].first.second = rk[bt - 1][i + (1 << (bt - 1))];
    }
    sort(tmp, tmp + n);
    int now = 0;
    for (int i = 1; i < n; i++) {
      int id = tmp[i].second;
      now += (tmp[i].first != tmp[i - 1].first);
      rk[bt][id] = now;
    }
  }
  sort(p, p + n, cmp);
  for (int i = 1; i < n; i++) l[i] = lcp(p[i], p[i - 1], n);
  g.push_back(n);
  long long now = 0, ans = 0;
  for (int i = n - 1; i > 0; i--) {
    while (g.back() != n && l[g.back()] > l[i]) {
      int last = g.back();
      g.pop_back();
      now -= 1ll * l[last] * (g.back() - last);
    }
    now += 1ll * (g.back() - i) * l[i];
    ans += now;
    g.push_back(i);
  }
  cout << ans + 1ll * (n + 1) * n / 2 << '\n';
}