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;
long long ans;
int n, diff[100005], anc[19][100005], rl[100005], pv[100005];
pair<long long, long long> nv[100005];
char s[100005];
int L[100005], R[100005], stk[100005], p;
int main() {
  cin.getline(s, 100005);
  n = strlen(s);
  for (int i = (0); i <= (17); i++) {
    if (i == 0)
      for (int j = (0); j <= (n - 1); j++) nv[j] = make_pair(s[j], j);
    else
      for (int j = (0); j <= (n - 1); j++) {
        if (j + (1 << i - 1) >= n)
          nv[j] = make_pair(1LL * anc[i - 1][j] * 100005, j);
        else
          nv[j] = make_pair(
              1LL * anc[i - 1][j] * 100005 + 1 + anc[i - 1][j + (1 << i - 1)],
              j);
      }
    sort(nv, nv + n);
    int rk = 0;
    anc[i][nv[0].second] = 0;
    for (int j = (1); j <= (n - 1); j++) {
      if (nv[j].first != nv[j - 1].first) rk++;
      anc[i][nv[j].second] = rk;
    }
  }
  for (int i = (0); i <= (n - 1); i++) rl[anc[17][i]] = i;
  for (int i = (1); i <= (n - 1); i++) {
    int lt = rl[i - 1], rt = rl[i];
    diff[i] = 0;
    for (int j = (17); j >= (0); j--) {
      if (anc[j][lt] == anc[j][rt]) {
        diff[i] += (1 << j);
        lt += (1 << j);
        rt += (1 << j);
        if (lt == n || rt == n) break;
      }
    }
  }
  ans = 1LL * n * (n + 1) / 2;
  p = 0;
  for (int i = (0); i <= (n - 1); i++) {
    while (p > 0 && diff[stk[p - 1]] >= diff[i]) p--;
    if (p == 0)
      L[i] = 0;
    else
      L[i] = stk[p - 1] + 1;
    stk[p++] = i;
  }
  p = 0;
  for (int i = (n - 1); i >= (0); i--) {
    while (p > 0 && diff[stk[p - 1]] >= diff[i]) p--;
    if (p == 0)
      R[i] = n - 1;
    else
      R[i] = stk[p - 1] - 1;
    stk[p++] = i;
  }
  memset(pv, 0, sizeof(pv));
  for (int i = (0); i <= (n - 1); i++) {
    ans =
        ans + 1LL * (R[i] - i + 1) * (i + 1 - max(L[i], pv[diff[i]])) * diff[i];
    pv[diff[i]] = i + 1;
  }
  printf("%lld\n", ans);
  return 0;
}