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 maxLen = 300000;
struct state {
  int len, pre, cnt;
  int next[26];
};
int sz, last;
state st[maxLen];
void clear() { sz = last = 1; }
void extend(int c) {
  int cur = ++sz, now;
  st[cur].len = st[last].len + 1;
  st[cur].cnt = 1;
  now = last;
  last = cur;
  for (; now && !st[now].next[c]; now = st[now].pre) st[now].next[c] = cur;
  if (!now)
    st[cur].pre = 1;
  else {
    int t = st[now].next[c];
    if (st[now].len + 1 == st[t].len)
      st[cur].pre = t;
    else {
      int clone = ++sz;
      st[clone].len = st[now].len + 1;
      memcpy(st[clone].next, st[t].next, sizeof(st[clone].next));
      st[clone].pre = st[t].pre;
      for (; now && st[now].next[c] == t; now = st[now].pre)
        st[now].next[c] = clone;
      st[t].pre = st[cur].pre = clone;
    }
  }
}
int k, l, now;
long long ans;
int rk[maxLen];
char ch[maxLen];
inline bool cmp(int x, int y) { return st[x].len > st[y].len; }
int main() {
  clear();
  scanf("%s", ch + 1);
  l = strlen(ch + 1);
  for (int i = 1; i <= l; i++) extend(ch[i] - 'a');
  for (int i = 2; i <= sz; i++) rk[i] = i;
  sort(rk + 2, rk + sz + 1, cmp);
  for (int i = 2; i <= sz; i++) {
    ans += (long long)(st[rk[i]].len - st[st[rk[i]].pre].len) * st[rk[i]].cnt *
           (st[rk[i]].cnt + 1) / 2;
    st[st[rk[i]].pre].cnt += st[rk[i]].cnt;
  }
  printf("%I64d\n", ans);
  return 0;
}