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 <class T>
inline void read(T &x) {
  x = 0;
  T f = 1;
  char ch = getchar();
  while (!(ch >= '0' && ch <= '9')) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
template <class T>
inline void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x / 10) write(x / 10);
  putchar('0' + x % 10);
}
template <class T>
inline void Max(T &x, T y) {
  if (y > x) x = y;
}
template <class T>
inline void Min(T &x, T y) {
  if (y < x) x = y;
}
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 10;
int ch[N][26], pre[N], mxl[N], last = 1, cnt = 1, len, sz[N];
char s[N];
inline void Extend(int c) {
  int p = last, np = ++cnt;
  last = np;
  mxl[np] = mxl[p] + 1;
  sz[np] = 1;
  for (; p && !ch[p][c]; p = pre[p]) ch[p][c] = np;
  if (!p) {
    pre[np] = 1;
    return;
  }
  int q = ch[p][c], nq = ++cnt;
  if (mxl[q] == mxl[p] + 1) {
    pre[np] = q;
    --cnt;
    return;
  }
  memcpy(ch[nq], ch[q], sizeof(ch[q]));
  mxl[nq] = mxl[p] + 1, pre[nq] = pre[q], pre[q] = pre[np] = nq;
  for (; ch[p][c] == q; p = pre[p]) ch[p][c] = nq;
}
int tt[N], qq[N];
long long ans;
inline void init() {
  for (int i = 1; i <= cnt; i++) tt[mxl[i]]++;
  for (int i = 1; i <= cnt; i++) tt[i] += tt[i - 1];
  for (int i = 1; i <= cnt; i++) qq[tt[mxl[i]]--] = i;
  for (int i = cnt; i >= 1; i--) sz[pre[qq[i]]] += sz[qq[i]];
  for (int i = 2; i <= cnt; i++)
    ans += 1ll * sz[i] * (sz[i] + 1) * (mxl[i] - mxl[pre[i]]) / 2;
  write(ans);
}
int main() {
  scanf("%s", s + 1);
  len = strlen(s + 1);
  for (int i = 1; i <= len; i++) Extend(s[i] - 'a');
  init();
  return 0;
}