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>
#pragma GCC optimize(2)
using namespace std;
inline int read() {
  char c = getchar();
  int x = 0;
  bool f = 0;
  for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
  return x;
}
char s[2000005];
int l, tot = 1, lst = 1;
int ch[2000005][26], len[2000005], fail[2000005];
int sz[2000005];
void append(int c) {
  int u = ++tot, p = lst;
  len[u] = len[p] + 1, sz[tot] = 1;
  for (; p && !ch[p][c]; p = fail[p]) ch[p][c] = u;
  if (!p)
    fail[u] = 1;
  else {
    int q = ch[p][c];
    if (len[q] == len[p] + 1)
      fail[u] = q;
    else {
      int cc = ++tot;
      fail[cc] = fail[q], len[cc] = len[p] + 1;
      memcpy(ch[cc], ch[q], sizeof(ch[cc]));
      for (; p && ch[p][c] == q; p = fail[p]) ch[p][c] = cc;
      fail[q] = fail[u] = cc;
    }
  }
  lst = u;
}
long long res;
struct edge {
  int to, nxt;
} e[2000005];
int cnt, head[2000005];
inline void adde(int u, int v) {
  e[++cnt] = (edge){v, head[u]};
  head[u] = cnt;
}
void dfs(int u) {
  for (int i = head[u]; i; i = e[i].nxt) {
    dfs(e[i].to);
    sz[u] += sz[e[i].to];
  }
  res += 1ll * sz[u] * (sz[u] + 1) / 2 * (len[u] - len[fail[u]]);
}
signed main() {
  cin >> s;
  l = strlen(s);
  for (register int i = (0); i <= (l - 1); ++i) append(s[i] - 'a');
  for (register int i = (2); i <= (tot); ++i) adde(fail[i], i);
  dfs(1);
  cout << res;
  return 0;
}