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;
const int N = 100010;
int n;
int sa[N], rk[N], cnt[N], oldrk[N << 1], id[N], px[N];
char s[N];
int ht[N];
pair<int, int> stk[N];
int tt;
inline bool cmp(int x, int y, int k) {
  return oldrk[x] == oldrk[y] && oldrk[x + k] == oldrk[y + k];
}
void SA() {
  int i, p = 0, k, m = 300;
  memset(cnt, 0, sizeof cnt);
  for (i = 1; i <= n; i++) cnt[rk[i] = s[i]]++;
  for (i = 1; i <= m; i++) cnt[i] += cnt[i - 1];
  for (i = n; i >= 1; i--) sa[cnt[rk[i]]--] = i;
  for (k = 1; k < n; k <<= 1, m = p) {
    for (p = 0, i = n; i > n - k; i--) id[++p] = i;
    for (i = 1; i <= n; i++) {
      if (sa[i] > k) id[++p] = sa[i] - k;
    }
    memset(cnt, 0, sizeof cnt);
    for (i = 1; i <= n; i++) cnt[px[i] = rk[id[i]]]++;
    for (i = 1; i <= m; i++) cnt[i] += cnt[i - 1];
    for (i = n; i >= 1; i--) sa[cnt[px[i]]--] = id[i];
    memcpy(oldrk, rk, sizeof rk);
    for (p = 0, i = 1; i <= n; i++) {
      rk[sa[i]] = cmp(sa[i], sa[i - 1], k) ? p : ++p;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (rk[i] == 1) continue;
    int k = max(ht[rk[i - 1]] - 1, 0);
    while (i + k <= n && s[i + k] == s[sa[rk[i] - 1] + k]) k++;
    ht[rk[i]] = k;
  }
}
long long f(long long x) { return x * (x + 1) / 2; }
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  SA();
  long long res = 0;
  for (int i = 1; i <= n; i++) {
    res += n - sa[i] + 1 - max(ht[i], ht[i + 1]);
  }
  for (int i = 1; i <= n + 1; i++) {
    int left = i;
    while (tt && stk[tt].first > ht[i]) {
      int num = i - stk[tt].second + 1;
      int len = stk[tt].first - max(ht[i], stk[tt - 1].first);
      res += f(num) * len;
      left = min(left, stk[tt].second);
      tt--;
    }
    stk[++tt] = {ht[i], left};
  }
  cout << res << endl;
  return 0;
}