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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC optimize("no-stack-protector,fast-math")
using namespace std;
const long long N = 1e5 + 1000, OO = 1e15 + 7, T = 13, M = 998244353, P = 6151,
                SQ = 360, LG = 18;
long long rnk[N][LG + 10], p[N], pw, n;
bool cmp(long long x, long long y) {
  if (rnk[x][pw - 1] != rnk[y][pw - 1]) return rnk[x][pw - 1] < rnk[y][pw - 1];
  if (max(x, y) + (1 << pw - 1) > n) return x > y;
  return rnk[x + (1 << pw - 1)][pw - 1] < rnk[y + (1 << pw - 1)][pw - 1];
}
void Suffix(string s) {
  n = (long long)s.size();
  for (long long i = 1; i <= n; i++) rnk[i][0] = s[i - 1] - 'a' + 1, p[i] = i;
  for (long long i = 1; i < LG; i++) {
    pw = i;
    sort(p + 1, p + n + 1, cmp);
    rnk[p[1]][i] = 1;
    for (long long j = 2; j <= n; j++)
      rnk[p[j]][i] = rnk[p[j - 1]][i] + cmp(p[j - 1], p[j]);
  }
}
long long Lcp(long long x, long long y) {
  long long cnt = 0;
  for (long long i = LG - 1; i >= 0; i--)
    if (rnk[x][i] == rnk[y][i]) x += (1 << i), y += (1 << i), cnt += (1 << i);
  return cnt;
}
long long l[N], r[N], lc[N];
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  string s;
  cin >> s;
  Suffix(s);
  long long ans = 0;
  l[1] = 1;
  for (long long i = 1; i < n; i++) {
    l[i] = i - 1;
    lc[i] = Lcp(p[i], p[i + 1]);
    while (l[i] && lc[l[i]] >= lc[i]) l[i] = l[l[i]];
  }
  r[n] = n;
  for (long long i = n - 1; i; i--) {
    r[i] = i + 1;
    while (r[i] != n && lc[r[i]] > lc[i]) r[i] = r[r[i]];
    ans += lc[i] * (r[i] - i) * (i - l[i]);
  }
  ans += (n * (n + 1)) / 2;
  cout << ans << endl;
  return 0;
}