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, k, a[100005], r[200005], sa[100005], V[100005], R[100005], x[100005],
    y[100005], z[100005], f[100005], h[100005];
vector<int> v[100005];
long long ans;
int main() {
  ios::sync_with_stdio(0);
  cin >> s;
  n = s.size();
  for (int i = 1; i <= n; ++i) r[i] = a[i] = s[i - 1];
  for (int _ = 1; _ < n; _ <<= 1) {
    k = 0;
    for (int i = 1; i <= n; ++i) v[r[i + _]].push_back(i);
    for (int i = 0; i <= 100000; ++i) {
      for (int j = 0; j < v[i].size(); ++j) V[++k] = v[i][j];
      v[i].clear();
    }
    k = 0;
    for (int i = 1; i <= n; ++i) v[r[V[i]]].push_back(V[i]);
    for (int i = 0; i <= 100000; ++i) {
      for (int j = 0; j < v[i].size(); ++j) V[++k] = v[i][j];
      v[i].clear();
    }
    k = 0;
    for (int i = 1; i <= n; ++i) {
      if (i == 1 || make_pair(r[V[i]], r[V[i] + _]) >
                        make_pair(r[V[i - 1]], r[V[i - 1] + _]))
        ++k;
      R[V[i]] = k;
    }
    for (int i = 1; i <= n; ++i) r[i] = R[i];
  }
  for (int i = 1; i <= n; ++i) sa[r[i]] = i;
  k = 0;
  for (int i = 1; i <= n; ++i) {
    if (r[i] == 1) {
      k = 0;
      continue;
    }
    int j = sa[r[i] - 1];
    if (k) --k;
    while (a[i + k] == a[j + k]) ++k;
    h[r[i] - 1] = k;
  }
  k = 0;
  ans = n * (n + 1ll) / 2ll;
  for (int i = 1; i <= n; ++i) {
    int o = i;
    while (k && x[k] > h[i])
      ans += min(y[k], x[k] - h[i]) * (i * 1ll - z[k]) * (1ll + i - z[k]) / 2ll,
          o = z[k--];
    if (x[k] < h[i]) x[++k] = h[i], y[k] = h[i] - x[k - 1], z[k] = o;
  }
  printf("%I64d\n", ans);
  return 0;
}