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 long long mod = 1e9 + 7;
struct node {
  int x;
  int y;
};
int n, m;
long long ans = 0, cnt = 0;
int sa[100000 + 5], rrank[100000 + 5], tax[100000 + 5], tp[100000 + 5],
    height[100000 + 5];
char s[100000 + 5];
stack<node> st;
bool cmp(int* f, int x, int y, int w) {
  return f[x] == f[y] && f[x + w] == f[y + w];
}
void ssort() {
  for (int i = 0; i <= m; i++) tax[i] = 0;
  for (int i = 1; i <= n; i++) tax[rrank[tp[i]]]++;
  for (int i = 1; i <= m; i++) tax[i] += tax[i - 1];
  for (int i = n; i >= 1; i--) sa[tax[rrank[tp[i]]]--] = tp[i];
}
void suffix() {
  for (int i = 1; i <= n; i++) rrank[i] = s[i], tp[i] = i;
  m = 127;
  ssort();
  for (int w = 1, p = 1, i; p < n; w += w, m = p) {
    for (p = 0, i = n - w + 1; i <= n; i++) tp[++p] = i;
    for (i = 1; i <= n; i++)
      if (sa[i] > w) tp[++p] = sa[i] - w;
    ssort();
    swap(rrank, tp);
    rrank[sa[1]] = p = 1;
    for (int i = 2; i <= n; i++)
      rrank[sa[i]] = cmp(tp, sa[i], sa[i - 1], w) ? p : ++p;
  }
  for (int i = 1, j = 0; i <= n; i++) {
    if (j) j--;
    while (s[i + j] == s[sa[rrank[i] - 1] + j]) j++;
    height[rrank[i]] = j;
  }
}
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  ans = (long long)n * (n + 1) / 2;
  suffix();
  for (int i = 1; i <= n; i++) {
    node now;
    now.x = height[i];
    now.y = 1;
    while (!st.empty() && st.top().x > now.x) {
      cnt -= (long long)st.top().x * st.top().y;
      now.y += st.top().y;
      st.pop();
    }
    cnt += (long long)now.x * now.y;
    st.push(now);
    ans += cnt;
  }
  cout << ans << endl;
  return 0;
}