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 maxN = 1000010;
string s;
int len;
int ra[maxN], tra[maxN], sa[maxN], tmpsa[maxN], cnt[maxN];
void countSort(int k) {
  int maxr = max(300, len);
  memset(cnt, 0, sizeof(cnt[0]) * (maxr + 5));
  for (int i = 0; i < len; i++) cnt[sa[i] + k < len ? ra[sa[i] + k] : 0]++;
  for (int i = 0, t, sum = 0; i < maxr; i++) t = cnt[i], cnt[i] = sum, sum += t;
  for (int i = 0; i < len; i++)
    tmpsa[cnt[sa[i] + k < len ? ra[sa[i] + k] : 0]++] = sa[i];
  for (int i = 0; i < len; i++) sa[i] = tmpsa[i];
}
void buildsa() {
  for (int i = 0; i < len; i++) ra[i] = s[i], sa[i] = i;
  for (int k = 1; k < len; k <<= 1) {
    countSort(k);
    countSort(0);
    int r = 0;
    tra[sa[0]] = 0;
    for (int i = 1; i < len; i++) {
      int t1 = (sa[i] + k < len ? ra[sa[i] + k] : 0);
      int t2 = (sa[i - 1] + k < len ? ra[sa[i - 1] + k] : 0);
      tra[sa[i]] = (ra[sa[i]] == ra[sa[i - 1]] && t1 == t2) ? r : ++r;
    }
    for (int i = 0; i < len; i++) ra[i] = tra[i];
    if (ra[sa[len - 1]] == len - 1) break;
  }
}
int phi[maxN], LCP[maxN];
void computeLCP() {
  phi[sa[0]] = -1;
  for (int i = 1; i < len; i++) phi[sa[i]] = sa[i - 1];
  int L = 0;
  for (int i = 0; i < len; i++) {
    if (phi[i] == -1) {
      LCP[ra[i]] = 0;
      continue;
    }
    while (L < len && s[i + L] == s[phi[i] + L]) L++;
    int TL = L;
    LCP[ra[i]] = TL;
    L = max(L - 1, 0);
  }
}
int tlft[maxN], trgt[maxN];
void hgram(int data[], int n) {
  data[0] = -1;
  data[n + 1] = -1;
  stack<int> bosta;
  bosta.push(0);
  for (int i = 1; i <= n + 1; i++) {
    while (data[i] < data[bosta.top()]) {
      trgt[bosta.top()] = i;
      bosta.pop();
    }
    bosta.push(i);
  }
  bosta.push(n + 1);
  for (int i = n + 1; i >= 0; i--) {
    while (data[i] < data[bosta.top()]) {
      tlft[bosta.top()] = i;
      bosta.pop();
    }
    bosta.push(i);
  }
}
int lim[maxN];
int main() {
  cin >> s;
  s += '&';
  len = s.size();
  buildsa();
  computeLCP();
  hgram(LCP, len - 1);
  long long tot = 0;
  for (int i = 0; i < len; i++) {
    long long unq = len - 1 - sa[i];
    long long del = LCP[i];
    if (i != len - 1) del = max(del, (long long)LCP[i + 1]);
    unq -= del;
    tot += unq;
    if (!LCP[i] || i < lim[LCP[i]]) continue;
    lim[LCP[i]] = max(lim[LCP[i]], trgt[i]);
    long long occ = trgt[i] - tlft[i];
    long long strcnt =
        min(LCP[i], min(LCP[i] - LCP[tlft[i]], LCP[i] - LCP[trgt[i]]));
    tot += (occ * (occ + 1) * strcnt) / 2;
  }
  cout << tot << endl;
}