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;
int n, m;
long long k;
bool debug = false;
struct SuffixArray {
  pair<int, int> suffix[100005];
  int sa[100005], rank[100005], height[100005], h[100005];
  void init(int n, const char a[100005]) {
    sa[0] = 0;
    for (int i = 0; i < n; ++i) {
      rank[i] = a[i];
    }
    for (int m = 1; m < n; m <<= 1) {
      for (int i = 0; i < n; ++i) {
        sa[i] = i;
        suffix[i] = {rank[i], i + m < n ? rank[i + m] : -1};
      }
      sort(sa, sa + n,
           [&](int lhs, int rhs) { return suffix[lhs] < suffix[rhs]; });
      for (int i = 0; i < n; ++i) {
        if (i == 0 || (suffix[sa[i]] != suffix[sa[i - 1]])) {
          rank[sa[i]] = i;
        } else {
          rank[sa[i]] = rank[sa[i - 1]];
        }
      }
    }
    for (int i = 0; i < n; ++i) {
      if (rank[i] == 0) {
        h[i] = 0;
      } else {
        int x = i, y = sa[rank[i] - 1], z = max(0, h[i - 1] - 1);
        while (x + z < n && y + z < n && a[x + z] == a[y + z]) {
          ++z;
        }
        h[i] = z;
      }
    }
    for (int i = 0; i < n; ++i) {
      height[i] = h[sa[i]];
    }
  }
} sa;
char s[100005];
long long cnt(long long x) { return x * (x + 1) / 2; }
int main() {
  scanf("%s", s);
  n = strlen(s);
  sa.init(n, s);
  vector<pair<pair<int, int>, int>> v;
  sa.height[n] = 0;
  sa.sa[n] = n;
  long long ans = 0;
  for (int i = 0; i <= n; i++) {
    while (!v.empty() && v.back().first.second > sa.height[i]) {
      auto &b = v.back();
      if (b.first.first < sa.height[i]) {
        ans += (b.first.second - sa.height[i]) * cnt(i - b.second);
        b.first.second = sa.height[i];
      } else {
        ans += (b.first.second - b.first.first) * cnt(i - b.second);
        v.pop_back();
      }
    }
    if (sa.height[i] < n - sa.sa[i]) {
      v.push_back({{sa.height[i], n - sa.sa[i]}, i});
    }
  }
  printf("%lld\n", ans);
  return 0;
}