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;
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
const long double PI = 2 * acos(0.0);
const long double eps = 1e-15;
const int infi = 1e9;
const long long Linfi = (long long)1e18;
const long long MOD = 1000000007;
const int c1 = 31;
const int c2 = 37;
string S;
int n, gap;
int sa[100005], pos[100005], tmp[100005], lcp[100005];
int a[100005], parent[100005], cnt[100005];
vector<pair<int, int> > edges[100005];
bool sufCmp(int i, int j) {
  if (pos[i] != pos[j]) return pos[i] < pos[j];
  i += gap;
  j += gap;
  return (i < n && j < n) ? pos[i] < pos[j] : i > j;
}
void buildSA() {
  n = S.size();
  for (int i = 0; i < n; i++) sa[i] = i, pos[i] = S[i];
  for (gap = 1;; gap *= 2) {
    sort(sa, sa + n, sufCmp);
    for (int i = 0; i < n - 1; i++)
      tmp[i + 1] = tmp[i] + sufCmp(sa[i], sa[i + 1]);
    for (int i = 0; i < n; i++) pos[sa[i]] = tmp[i];
    if (tmp[n - 1] == n - 1) break;
  }
}
void buildLCP() {
  for (int i = 0, k = 0; i < n; ++i)
    if (pos[i] != n - 1) {
      for (int j = sa[pos[i] + 1]; S[i + k] == S[j + k];) ++k;
      lcp[pos[i]] = k;
      if (k) --k;
    }
}
int find_set(int first) {
  return (parent[first] == first ? first
                                 : parent[first] = find_set(parent[first]));
}
void Union(int first, int second) {
  int xx = find_set(first), yy = find_set(second);
  parent[xx] = yy;
  cnt[yy] += cnt[xx];
}
void solve() {
  memset(sa, 0, sizeof(sa));
  memset(lcp, 0, sizeof(lcp));
  buildSA();
  buildLCP();
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    edges[lcp[i]].push_back(pair<int, int>(i, i + 1));
  }
  for (int i = 0; i <= n; i++) parent[i] = i, cnt[i] = 1;
  for (int i = n; i >= 1; i--) {
    for (auto e : edges[i]) {
      int u = e.first, v = e.second;
      ans += 1LL * cnt[find_set(u)] * cnt[find_set(v)] * i;
      Union(u, v);
    }
  }
  ans += 1LL * n * (n + 1) / 2;
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> S;
  solve();
  return 0;
}