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 = 123456;
char s[N];
int sa[N];
int group[N], new_group[N], ptr_group[N];
int stk[N];
int lcp[N];
int p[N], sz[N];
vector<int> cand[N];
int get(int x) { return p[x] == -1 ? x : (p[x] = get(p[x])); }
bool cmp(int i, int j) { return s[i] < s[j]; }
int main() {
  scanf("%s", s);
  int n = strlen(s);
  for (int i = 0; i < n; i++) {
    sa[i] = i;
  }
  sort(sa, sa + n, cmp);
  group[sa[0]] = 0;
  for (int i = 1; i < n; i++) {
    group[sa[i]] = group[sa[i - 1]];
    if (s[sa[i - 1]] != s[sa[i]]) {
      group[sa[i]]++;
    }
  }
  for (int step = 1; group[sa[n - 1]] != n - 1; step <<= 1) {
    int ptr = 0;
    for (int i = n - step; i < n; i++) {
      stk[ptr++] = i;
    }
    for (int i = 0; i < n; i++) {
      if (sa[i] >= step) {
        stk[ptr++] = sa[i] - step;
      }
    }
    for (int i = n - 1; i >= 0; i--) {
      ptr_group[group[sa[i]]] = i;
    }
    for (int i = 0; i < n; i++) {
      int x = stk[i];
      sa[ptr_group[group[x]]++] = x;
    }
    new_group[sa[0]] = 0;
    for (int i = 1; i < n; i++) {
      if (group[sa[i]] != group[sa[i - 1]]) {
        new_group[sa[i]] = new_group[sa[i - 1]] + 1;
      } else {
        int pre = (sa[i - 1] + step < n ? group[sa[i - 1] + step] : -1);
        int cur = (sa[i] + step < n ? group[sa[i] + step] : -1);
        new_group[sa[i]] = new_group[sa[i - 1]] + (pre != cur);
      }
    }
    copy(new_group, new_group + n, group);
  }
  int k = 0;
  for (int i = 0; i < n; i++) {
    if (group[i] == n - 1) {
      k = 0;
    } else {
      k = max(k - 1, 0);
      int j = sa[group[i] + 1];
      while (i + k < n && j + k < n && s[i + k] == s[j + k]) {
        k++;
      }
      lcp[group[i]] = k;
    }
  }
  for (int i = 0; i < n; i++) {
    p[i] = -1;
    sz[i] = 1;
  }
  for (int i = 0; i < n - 1; i++) {
    cand[lcp[i]].push_back(i);
  }
  long long acc = 0;
  long long ans = 1LL * n * (n + 1) / 2;
  for (int h = n - 1; h >= 1; h--) {
    for (int j = 0; j < (int)cand[h].size(); j++) {
      int i = cand[h][j];
      int x = get(i);
      int y = get(i + 1);
      if (x != y) {
        p[x] = y;
        acc += 1LL * sz[x] * sz[y];
        sz[y] += sz[x];
      }
    }
    ans += acc;
  }
  printf("%lld\n", ans);
  return 0;
}