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 ch[200005][26];
int fa[200005], cnt, last, rt;
long long sz[200005], dis[200005];
void init() {
  cnt = 0;
  rt = last = ++cnt;
  memset(fa, 0, sizeof(fa));
  memset(dis, 0, sizeof(dis));
  memset(sz, 0, sizeof(sz));
  memset(ch, 0, sizeof(ch));
}
inline int newnode(int x) {
  dis[++cnt] = x;
  return cnt;
}
void add(int x) {
  int np = newnode(dis[last] + 1), p = last;
  for (; p && (!ch[p][x]); p = fa[p]) ch[p][x] = np;
  if (!p)
    fa[np] = rt;
  else {
    int q = ch[p][x];
    if (dis[q] == dis[p] + 1)
      fa[np] = q;
    else {
      int nq = newnode(dis[p] + 1);
      memcpy(ch[nq], ch[q], sizeof(ch[q]));
      fa[nq] = fa[q];
      fa[q] = fa[np] = nq;
      for (; ch[p][x] == q; p = fa[p]) ch[p][x] = nq;
    }
  }
  last = np;
  sz[np] = 1;
}
char s[200005];
int A, B;
int c[200005], t[200005];
long long solve(long long n) { return n * (n + 1) / 2; }
long long count() {
  memset(c, 0, sizeof(c));
  memset(t, 0, sizeof(t));
  long long ans = 0;
  for (int i = 1; i <= cnt; i++) c[dis[i]]++;
  for (int i = 1; i <= cnt; i++) c[i] += c[i - 1];
  for (int i = cnt; i; i--) t[c[dis[i]]--] = i;
  for (int i = cnt; i; i--) {
    int x = t[i];
    sz[fa[x]] += sz[x];
  }
  for (int i = 2; i <= cnt; ++i)
    ans += 1ll * solve(sz[i]) * (dis[i] - dis[fa[i]]);
  return ans;
}
int main() {
  while (~scanf("%s", s + 1)) {
    init();
    for (int i = 1; s[i]; ++i) add(s[i] - 'a');
    cout << count() << endl;
  }
  return 0;
}