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 MN = 100100, LG = 19;
char N[MN];
int kmr[LG][MN], LCP[MN], suffArr[MN], rudy[MN];
long long dp[MN];
vector<pair<int, int> > S;
pair<int, pair<int, int> > cand[MN], tmp[MN];
vector<int> kub[MN];
int mav, n;
void radS() {
  int pos = 0;
  for (int i = 1; i <= n; ++i) kub[cand[i].second.second].push_back(i);
  for (int i = 0; i <= mav; ++i) {
    int s = kub[i].size();
    for (int j = 0; j < s; ++j) tmp[++pos] = cand[kub[i][j]];
  }
  pos = 0;
  for (int i = 0; i <= mav; ++i) kub[i].clear();
  for (int i = 1; i <= n; ++i) kub[tmp[i].second.first].push_back(i);
  for (int i = 0; i <= mav; ++i) {
    int s = kub[i].size();
    for (int j = 0; j < s; ++j) cand[++pos] = tmp[kub[i][j]];
  }
  for (int i = 0; i <= mav; ++i) kub[i].clear();
}
int main() {
  scanf("%s", N);
  n = strlen(N);
  mav = max(203, n + 7);
  for (int i = n; i > 0; --i) N[i] = N[i - 1];
  N[0] = '.';
  N[n + 1] = 200;
  for (int i = 1; i <= n; ++i) kmr[0][i] = N[i] - 'a' + 1;
  for (int i = 1; i < LG; ++i) {
    int p2 = (1 << (i - 1));
    for (int j = 1; j <= n; ++j) {
      if (j + p2 > n)
        cand[j] = make_pair(j, make_pair(kmr[i - 1][j], max(200, n + 5)));
      else
        cand[j] = make_pair(j, make_pair(kmr[i - 1][j], kmr[i - 1][j + p2]));
    }
    radS();
    int pos = 0;
    for (int j = 1; j <= n; ++j) {
      if (j == 1 || cand[j].second != cand[j - 1].second) ++pos;
      kmr[i][cand[j].first] = pos;
    }
  }
  for (int i = 1; i <= n; ++i) suffArr[kmr[LG - 1][i]] = i;
  for (int i = 1; i <= n; ++i) rudy[suffArr[i]] = i;
  int currl = 0;
  for (int i = 1; i <= n; ++i) {
    while (N[i + currl] == N[suffArr[rudy[i] - 1] + currl]) ++currl;
    LCP[rudy[i]] = currl;
    if (currl) --currl;
  }
  long long res = n;
  res *= ((long long)(n + 1));
  res >>= 1;
  for (int i = 1; i <= n; ++i) res -= LCP[i];
  S.push_back(make_pair(-1, 1));
  for (int i = 1; i <= n; ++i) {
    while (S.back().first >= LCP[i]) S.pop_back();
    long long cc = LCP[i];
    cc *= ((long long)(i - S.back().second + 1));
    dp[i] += cc;
    if (i > 1) {
      dp[i] += dp[S.back().second];
      if (dp[S.back().second]) dp[i] -= S.back().first;
      res += dp[i];
    }
    S.push_back(make_pair(LCP[i], i));
  }
  printf("%I64d", res);
}