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 = 100000 + 5;
int wa[N], wb[N], wc[N], wv[N];
bool 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 *x = wa, *y = wb;
  for (int i = 0; i < m; i++) wc[i] = 0;
  for (int i = 0; i < n; i++) wc[x[i] = r[i]]++;
  for (int i = 1; i < m; i++) wc[i] += wc[i - 1];
  for (int i = n - 1; i >= 0; i--) sa[--wc[x[i]]] = i;
  for (int j = 1, p = 1; p < n; j <<= 1, m = p) {
    p = 0;
    for (int i = n - j; i < n; i++) y[p++] = i;
    for (int i = 0; i < n; i++)
      if (sa[i] >= j) y[p++] = sa[i] - j;
    for (int i = 0; i < n; i++) wv[i] = x[y[i]];
    for (int i = 0; i < m; i++) wc[i] = 0;
    for (int i = 0; i < n; i++) wc[wv[i]]++;
    for (int i = 1; i < m; i++) wc[i] += wc[i - 1];
    for (int i = n - 1; i >= 0; i--) sa[--wc[wv[i]]] = y[i];
    int *t = x;
    x = y;
    y = t;
    p = 1;
    x[sa[0]] = 0;
    for (int i = 1; i < n; i++)
      x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;
  }
}
void calheight(int *r, int *sa, int *rk, int *height, int n) {
  for (int i = 1; i <= n; i++) rk[sa[i]] = i;
  int k = 0;
  for (int i = 0; i < n; height[rk[i++]] = k) {
    if (k) k--;
    for (int j = sa[rk[i] - 1]; r[i + k] == r[j + k]; k++)
      ;
  }
}
int a[N], sa[N], rk[N], height[N];
int b[N], h[N];
long long sum[N];
char s[N];
long long solve() {
  int n = strlen(s);
  for (int i = 0; i < n; i++) a[i] = s[i] - 'a' + 1;
  a[n] = 0;
  da(a, sa, n + 1, 28);
  calheight(a, sa, rk, height, n);
  int top = 0;
  long long ans = 1ll * n * (n + 1) / 2;
  for (int i = 1; i <= n; i++) {
    int cnt = 1;
    while (top > 0 && h[top] >= height[i]) {
      cnt += b[top];
      top--;
    }
    top++;
    h[top] = height[i];
    b[top] = cnt;
    sum[top] = sum[top - 1] + 1ll * h[top] * b[top];
    ans += sum[top];
  }
  return ans;
}
int main() {
  scanf("%s", s);
  printf("%I64d\n", solve());
  return 0;
}