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;
int wa[200010], wb[200010], wv[200010], wss[200010];
int cmp(int *r, int a, int b, int len) {
  return r[a] == r[b] && r[a + len] == r[b + len];
}
void da(int *r, int *sa, int n, int m) {
  int i, j, p, *x = wa, *y = wb, *t;
  for (i = 0; i < m; i++) wss[i] = 0;
  for (i = 0; i < n; i++) wss[x[i] = r[i]]++;
  for (i = 1; i < m; i++) wss[i] += wss[i - 1];
  for (i = n - 1; i >= 0; i--) sa[--wss[x[i]]] = i;
  for (j = p = 1; p < n; j *= 2, m = p) {
    for (p = 0, i = n - j; i < n; i++) y[p++] = i;
    for (i = 0; i < n; i++)
      if (sa[i] >= j) y[p++] = sa[i] - j;
    for (i = 0; i < n; i++) wv[i] = x[y[i]];
    for (i = 0; i < m; i++) wss[i] = 0;
    for (i = 0; i < n; i++) wss[wv[i]]++;
    for (i = 1; i < m; i++) wss[i] += wss[i - 1];
    for (i = n - 1; i >= 0; i--) sa[--wss[wv[i]]] = y[i];
    for (t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; i++)
      x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;
  }
}
int rak[200010], height[200010];
void calheight(int *r, int *sa, int n) {
  int i, j, k = 0;
  for (i = 0; i < n; i++) rak[sa[i]] = i;
  for (i = 0; i < n; height[rak[i++]] = k)
    for (k ? k-- : 0, j = sa[rak[i] - 1]; r[i + k] == r[j + k]; k++)
      ;
}
int r[200010];
char s[200010];
int sa[200010];
int f[200010][25];
int bit[200010];
void init_rmq(int n) {
  int i, j, temp;
  for (i = 1; i <= n; i++) f[i][0] = height[i];
  for (i = 1; (1 << i) <= n; i++) {
    for (j = 1; j <= n; j++) {
      temp = j + (1 << (i - 1));
      if (temp <= n)
        f[j][i] = min(f[j][i - 1], f[temp][i - 1]);
      else
        f[j][i] = f[j][i - 1];
    }
  }
  bit[1] = 0;
  for (i = 1; (1 << i) < 200010; i++)
    for (j = (1 << i); j < (1 << (i + 1)) && j < 200010; j++)
      bit[j] = bit[1 << (i - 1)] + 1;
}
int rmq_min(int left, int right) {
  int m = bit[right - left + 1];
  return min(f[left][m], f[right - (1 << m) + 1][m]);
}
int main() {
  memset(r, 0, sizeof(r));
  memset(sa, 0, sizeof(sa));
  memset(rak, 0, sizeof(rak));
  memset(height, 0, sizeof(height));
  scanf("%s", s);
  int n = 0;
  for (int i = 0; s[i]; i++) r[n++] = s[i];
  n++;
  da(r, sa, n, 256);
  calheight(r, sa, n);
  bool flag = false;
  int m = strlen(s);
  init_rmq(n);
  long long sum = 0;
  for (int i = 1; i < n && !flag; i++)
    if (sa[i] < m) {
      int rt = n;
      for (int j = height[i] + 1; j <= height[i + 1]; j++) {
        int lt = i + 1, mid, ans;
        while (lt <= rt) {
          mid = (lt + rt) / 2;
          if (rmq_min(i + 1, mid) >= j) {
            ans = mid;
            lt = mid + 1;
          } else
            rt = mid - 1;
        }
        sum += 1LL * (ans - i + 1) * (ans - i + 2) / 2;
      }
      sum += (m - sa[i] - max(height[i + 1], height[i]));
    }
  cout << sum << endl;
  return 0;
}