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 = 1e6 + 5;
const int mod = 1e9 + 7;
const int MOD = mod - 1;
const int inf = 0x3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-10;
int sa[N];
int rk[N];
int tmp[N];
int lcp[N];
int n, k;
bool cmp(int i, int j) {
  if (rk[i] != rk[j])
    return rk[i] < rk[j];
  else {
    int ri = i + k <= n ? rk[i + k] : -1;
    int rj = j + k <= n ? rk[j + k] : -1;
    return ri < rj;
  }
}
void bulid(string s, int *sa) {
  n = (int)s.size();
  for (int i = 0; i <= n; i++) {
    sa[i] = i;
    rk[i] = i < n ? s[i] : -1;
  }
  for (k = 1; k <= n; k *= 2) {
    sort(sa, sa + n + 1, cmp);
    tmp[sa[0]] = 0;
    for (int i = 1; i <= n; i++) {
      tmp[sa[i]] = tmp[sa[i - 1]] + (cmp(sa[i - 1], sa[i]) ? 1 : 0);
    }
    for (int i = 0; i <= n; i++) {
      rk[i] = tmp[i];
    }
  }
}
void LCP(string s, int *sa, int *lcp) {
  n = (int)s.size();
  for (int i = 0; i <= n; i++) rk[sa[i]] = i;
  int h = 0;
  lcp[0] = 0;
  for (int i = 0; i < n; i++) {
    int j = sa[rk[i] - 1];
    for (h ? h-- : 0; j + h < n && i + h < n && s[j + h] == s[i + h]; h++)
      ;
    lcp[rk[i] - 1] = h;
  }
}
stack<pair<int, int> > sta;
int main() {
  string s;
  cin >> s;
  n = s.length();
  bulid(s, sa);
  LCP(s, sa, lcp);
  long long ans = (long long)n * (long long)(n + 1) / 2;
  long long cnt = 0;
  for (int i = 0; i <= n; i++) {
    pair<int, int> ins = make_pair(lcp[i], 1);
    while (!sta.empty() && sta.top().first > ins.first) {
      cnt -= (long long)sta.top().first * sta.top().second;
      ins.second += sta.top().second;
      sta.pop();
    }
    cnt += (long long)ins.first * ins.second;
    sta.push(ins);
    ans += cnt;
  }
  cout << ans << endl;
  return 0;
}