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 M = 100010;
char s[M];
int n;
int wa[M], wb[M], wv[M], wc[M];
int r[M], sa[M], rak[M], height[M], dp[M][22], SIGMA = 0;
int logg[M];
long long cnt[M];
int cmp(int *r, int a, int b, int l) {
  return r[a] == r[b] && r[a + l] == r[b + l];
}
void da(int *r, int *sa, int n, int m) {
  int i, j, p, *first = wa, *second = wb, *t;
  for (i = 0; i < m; i++) wc[i] = 0;
  for (i = 0; i < n; i++) wc[first[i] = r[i]]++;
  for (i = 1; i < m; i++) wc[i] += wc[i - 1];
  for (i = n - 1; i >= 0; i--) sa[--wc[first[i]]] = i;
  for (j = 1, p = 1; p < n; j *= 2, m = p) {
    for (p = 0, i = n - j; i < n; i++) second[p++] = i;
    for (i = 0; i < n; i++)
      if (sa[i] >= j) second[p++] = sa[i] - j;
    for (i = 0; i < n; i++) wv[i] = first[second[i]];
    for (i = 0; i < m; i++) wc[i] = 0;
    for (i = 0; i < n; i++) wc[wv[i]]++;
    for (i = 1; i < m; i++) wc[i] += wc[i - 1];
    for (i = n - 1; i >= 0; i--) sa[--wc[wv[i]]] = second[i];
    for (t = first, first = second, second = t, p = 1, first[sa[0]] = 0, i = 1;
         i < n; i++)
      first[sa[i]] = cmp(second, sa[i - 1], sa[i], j) ? p - 1 : p++;
  }
}
void calheight(int *r, int *sa, int n) {
  int i, j, k = 0;
  for (i = 1; i <= n; i++) rak[sa[i]] = i;
  for (i = 0; i < n; height[rak[i++]] = k) {
    for (k ? k-- : 0, j = sa[rak[i] - 1]; r[i + k] == r[j + k]; k++)
      ;
  }
}
void initRMQ(int n) {
  for (int i = 1; i <= n; i++) dp[i][0] = height[i];
  for (int j = 1; (1 << j) <= n; j++) {
    for (int i = 1; i + (1 << j) - 1 <= n; i++) {
      dp[i][j] = min(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
    }
  }
  for (int i = 0; i < M; i++) {
    logg[i] = 0;
    while ((1 << (logg[i] + 1)) <= i) logg[i]++;
    ;
  }
}
int askRMQ(int L, int R) {
  int k = logg[R - L + 1];
  return min(dp[L][k], dp[R - (1 << k) + 1][k]);
}
void solve_for(int pos, int done) {
  int l = pos + 1, h = n;
  if (pos == n || height[pos + 1] <= done) {
    cnt[1] += n - sa[pos] - done;
    return;
  }
  while (l < h) {
    int m = (l + h + 1) / 2;
    if (askRMQ(pos + 1, m) > done)
      l = m;
    else
      h = m - 1;
  }
  cnt[l - pos + 1] += askRMQ(pos + 1, l) - done;
  solve_for(pos, askRMQ(pos + 1, l));
}
int main() {
  ios_base::sync_with_stdio(false);
  int i, j, k;
  scanf("%s", s);
  n = strlen(s);
  for (int i = 0; i < n; i++) {
    r[i] = s[i] - 'a' + 1;
    SIGMA = max(SIGMA, r[i]);
  }
  r[n] = 0;
  da(r, sa, n + 1, SIGMA + 1);
  calheight(r, sa, n);
  initRMQ(n);
  for (i = 1; i <= n; i++) solve_for(i, height[i]);
  long long sum = 0;
  for (long long i = 0; i <= n; i++) sum += (cnt[i]) * ((i * (i + 1)) / 2);
  printf("%lld\n", sum);
  return 0;
}