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_Automaton {
  struct state {
    int len, link, siz;
    int next[26];
  } t[101000 << 1];
  int cnt, lst;
  void init() { cnt = lst = 1; }
  void extend(int c) {
    int cur = ++cnt;
    t[cur].len = t[lst].len + 1, t[cur].siz = 1;
    int u = lst;
    while (u && !t[u].next[c]) t[u].next[c] = cur, u = t[u].link;
    if (u == 0)
      t[cur].link = 1;
    else {
      int q = t[u].next[c];
      if (t[q].len == t[u].len + 1)
        t[cur].link = q;
      else {
        int nq = ++cnt;
        t[nq].len = t[u].len + 1;
        memcpy(t[nq].next, t[q].next, sizeof(t[nq].next));
        t[nq].link = t[q].link;
        t[q].link = t[cur].link = nq;
        while (u && t[u].next[c] == q) t[u].next[c] = nq, u = t[u].link;
      }
    }
    lst = cur;
  }
} SAM;
int head[2 * 101000], to[2 * 101000], nxt[2 * 101000];
int cnt;
string s;
long long ans;
void init() { memset(head, -1, sizeof(head)), cnt = -1; }
void add_e(int a, int b) { nxt[++cnt] = head[a], head[a] = cnt, to[cnt] = b; }
void dfs(int x) {
  for (int i = head[x]; ~i; i = nxt[i])
    dfs(to[i]), SAM.t[x].siz += SAM.t[to[i]].siz;
  ans += (long long)(SAM.t[x].len - SAM.t[SAM.t[x].link].len) * SAM.t[x].siz *
         (SAM.t[x].siz + 1) / 2;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> s;
  SAM.init();
  for (char c : s) SAM.extend(c - 'a');
  init();
  for (int i = (2); i <= (SAM.cnt); i++) add_e(SAM.t[i].link, i);
  dfs(1);
  cout << ans << endl;
  return 0;
}