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;
struct suffix {
  int pos, first, second;
};
struct suffix_array {
 public:
  string S;
  vector<int> E[2 * 100005];
  vector<int> P[100005];
  int h[2 * 100005], par[2 * 100005], l[100005], pos[100005], n, dif, cur,
      Child[2 * 100005];
  long long Sum[2 * 100005];
  suffix a[100005];
  suffix b[100005];
  void Solve();
  void Dfs(int u);
  void Finish(int u);
  long long ans = 0;
} a;
void suffix_array::Finish(int u) {
  for (auto v : E[u]) {
    ans = ans + 1ll * Child[v] * (Child[v] + 1) * (h[v] - h[u]) / 2;
    Finish(v);
  }
}
void suffix_array::Dfs(int u) {
  for (auto v : E[u]) {
    Dfs(v);
    Child[u] += Child[v];
  }
}
void suffix_array::Solve() {
  int n = S.length();
  for (int i = 0; i < n; i++) {
    a[i] = {i, S[i] - '@', (i + 1 < n) ? (S[i + 1] - '@') : (-1)};
  }
  sort(a, a + n, [](const suffix& a, const suffix& b) {
    return (a.first < b.first || (a.first == b.first && a.second < b.second));
  });
  for (int i = 0; i < n; i++) {
    if (i && (a[i].first != a[i - 1].first || a[i].second != a[i - 1].second))
      ++dif;
    pos[a[i].pos] = dif;
  }
  int dist = 2;
  while (dist < n && dif < n - 1) {
    cur = 0;
    for (int i = 0; i < n; i++) {
      a[i].first = pos[a[i].pos];
      a[i].second = (a[i].pos + dist < n) ? pos[a[i].pos + dist] : (-1);
      if (a[i].second == -1) {
        b[cur++] = a[i];
      } else {
        P[a[i].second].push_back(i);
      }
    }
    for (int i = 0; i <= dif; i++) {
      for (auto v : P[i]) {
        b[cur++] = a[v];
      }
      P[i].clear();
    }
    cur = 0;
    for (int i = 0; i < n; i++) {
      P[b[i].first].push_back(i);
    }
    for (int i = 0; i <= dif; i++) {
      for (auto v : P[i]) {
        a[cur++] = b[v];
      }
      P[i].clear();
    }
    dif = 0;
    for (int i = 0; i < n; i++) {
      if (i && (a[i].first != a[i - 1].first || a[i].second != a[i - 1].second))
        ++dif;
      pos[a[i].pos] = dif;
    }
    dist = dist * 2;
  }
  int p = 0;
  for (int i = 0; i < n - 1; i++) {
    int j = a[pos[i] - 1].pos;
    while (S[i + p] == S[j + p]) ++p;
    l[pos[i]] = p;
    p--;
    if (p < 0) p = 0;
  }
  int cur = 0;
  int point = 0;
  for (int i = 1; i < n; i++) {
    while (h[cur] > l[i]) {
      cur = par[cur];
    }
    if (h[cur] == l[i]) {
      E[cur].push_back(++point);
      par[point] = cur;
      cur = point;
      h[point] = n - (a[i].pos + 1);
    } else {
      par[E[cur].back()] = point + 1;
      E[++point].push_back(E[cur].back());
      E[cur][E[cur].size() - 1] = point;
      h[point] = l[i];
      par[point] = cur;
      cur = point;
      E[cur].push_back(++point);
      par[point] = cur;
      cur = point;
      h[point] = n - (a[i].pos + 1);
    }
    Child[cur]++;
  }
  Sum[0] = 0;
  for (int i = 1; i <= n; i++) {
    Sum[i] = Sum[i - 1] + 1ll * i * (i + 1) / 2;
  }
}
int main() {
  cin >> a.S;
  a.S = a.S + '@';
  a.Solve();
  a.Dfs(0);
  a.Finish(0);
  cout << a.ans;
}