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 MAX_N = 100100;
struct SuffixArray {
  int sa[MAX_N], height[MAX_N];
  void build(const string &s) { build(s.c_str(), s.size()); }
  void build(const char *t, int n) {
    static int g[MAX_N], b[MAX_N];
    for (int i = 0; i < n + 1; ++i) {
      sa[i] = i, g[i] = t[i];
    }
    b[0] = b[n] = 0;
    sort(sa, sa + n + 1, SAComp(0, g));
    for (int h = 1; b[n] != n; h *= 2) {
      SAComp comp(h, g);
      sort(sa, sa + n + 1, comp);
      for (int i = 0; i < n; ++i) {
        b[i + 1] = b[i] + comp(sa[i], sa[i + 1]);
      }
      for (int i = 0; i < n + 1; ++i) {
        g[sa[i]] = b[i];
      }
    }
    for (int i = 0; i < n + 1; ++i) {
      b[sa[i]] = i;
    }
    int h = 0;
    for (int i = 0; i < n; ++i) {
      const int j = sa[b[i] - 1];
      while (j + h < n && i + h < n && t[j + h] == t[i + h]) h++;
      height[b[i]] = h;
      if (h) h--;
    }
    height[0] = -1;
  }
  struct SAComp {
    const int h, *g;
    SAComp(int h, int *g) : h(h), g(g) {}
    bool operator()(int l, int r) {
      return g[l] != g[r] ? g[l] < g[r] : g[l + h] < g[r + h];
    }
  };
} sa;
char buf[MAX_N];
long long cnt(long long n) { return n * (n + 1) / 2; }
int main() {
  int n;
  long long ans;
  vector<pair<pair<int, int>, int>> v;
  scanf("%s", buf);
  n = (int)strlen(buf);
  sa.build(buf, n);
  sa.height[n + 1] = 0;
  sa.sa[n + 1] = n;
  ans = 0;
  for (int i = 1; i <= n + 1; ++i) {
    while (!v.empty() && v.back().first.second > sa.height[i]) {
      auto &end = v.back();
      if (end.first.first < sa.height[i]) {
        ans += (end.first.second - sa.height[i]) * cnt(i - end.second);
        end.first.second = sa.height[i];
      } else {
        ans += (end.first.second - end.first.first) * cnt(i - end.second);
        v.pop_back();
      }
    }
    v.push_back({{sa.height[i], n - sa.sa[i]}, i});
  }
  printf("%lld\n", ans);
  return 0;
}