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;
template <typename T>
void cmax(T& x, T y) {
  if (x < y) x = y;
}
template <typename T>
void cmin(T& x, T y) {
  if (x > y) x = y;
}
inline int read() {
  int f = 1, x = 0;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return f * x;
}
const int N = 500005;
int n, al, tot, lk[N], go[N][26], len[N], siz[N];
long long ans;
char s[N];
vector<int> G[N];
inline void add(int c) {
  int cs = ++tot, t = al;
  al = cs, len[cs] = len[t] + 1, siz[cs] = 1;
  for (; t && !go[t][c]; t = lk[t]) go[t][c] = cs;
  if (!t) return lk[cs] = 1, void();
  int k = go[t][c];
  if (len[k] == len[t] + 1) return lk[cs] = k, void();
  int p = ++tot;
  lk[p] = lk[k], lk[cs] = lk[k] = p, len[p] = len[t] + 1;
  for (int i = (0), iend = (25); i <= iend; i++) go[p][i] = go[k][i];
  for (; t && go[t][c] == k; t = lk[t]) go[t][c] = p;
}
inline void dfs(int u) {
  for (auto v : G[u]) dfs(v), siz[u] += siz[v];
  ans += 1ll * (len[u] - len[lk[u]]) * siz[u] * (siz[u] + 1) / 2;
}
int main() {
  scanf("%s", s + 1), n = strlen(s + 1), al = tot = 1;
  for (int i = (1), iend = (n); i <= iend; i++) add(s[i] - 'a');
  for (int i = (2), iend = (tot); i <= iend; i++) G[lk[i]].push_back(i);
  dfs(1);
  printf("%lld", ans);
}