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>
int head[100000], succ[100000], succRank[100000];
int sa[100000], rank[100000];
int letter[128];
void da(char* str, int len) {
  memset(letter, 0, sizeof(letter));
  for (int i = 0; i < len; ++i)
    if (!letter[str[i]]) letter[str[i]] = 1;
  int total = -1;
  for (int i = 0; i < 128; ++i)
    if (letter[i]) letter[i] = ++total;
  memset(head, 255, sizeof(head));
  for (int i = len - 1; i >= 0; --i) {
    rank[i] = letter[str[i]];
    succ[i] = head[rank[i]];
    head[rank[i]] = i;
  }
  int j = 0;
  for (int i = 0; i < len; ++i) {
    while (head[j] == -1) ++j;
    sa[i] = head[j];
    head[j] = succ[head[j]];
  }
  for (int k = 1; k < len; k <<= 1) {
    for (int i = len - 1; i >= 0; --i)
      if (sa[i] - k >= 0) {
        succRank[sa[i] - k] = rank[sa[i]];
        succ[sa[i] - k] = head[rank[sa[i] - k]];
        head[rank[sa[i] - k]] = sa[i] - k;
      }
    for (int i = len - 1; i >= len - k; --i) {
      succRank[i] = -1;
      succ[i] = head[rank[i]];
      head[rank[i]] = i;
    }
    j = 0;
    total = -1;
    int preSuccRank = 0, preRank = 0;
    for (int i = 0; i < len; ++i) {
      while (head[j] == -1) ++j;
      sa[i] = head[j];
      if (i == 0 || preRank != rank[sa[i]] || preSuccRank != succRank[sa[i]]) {
        preRank = rank[sa[i]];
        rank[sa[i]] = ++total;
      } else {
        preRank = rank[sa[i]];
        rank[sa[i]] = total;
      }
      preSuccRank = succRank[sa[i]];
      head[j] = succ[head[j]];
    }
  }
}
int height[100000];
void calcHeight(char* str, int len) {
  int i, j, k = 0;
  for (i = 0; i < len; ++i) {
    if (k != 0) --k;
    if (rank[i] == 0) continue;
    j = sa[rank[i] - 1];
    while (str[i + k] == str[j + k]) ++k;
    height[rank[i]] = k;
  }
}
char str[100000 + 2];
int main() {
  using namespace std;
  while (gets(str)) {
    int len = strlen(str);
    da(str, len);
    calcHeight(str, len);
    stack<int> s;
    s.push(0);
    long long ans = 0;
    int k;
    for (int i = 1; i < len; ++i) {
      ans += len - sa[i - 1] - max(height[i - 1], height[i]);
      k = i;
      while (s.size() > 1 && height[s.top()] > height[i]) {
        int b = i - (s.top() - 1);
        k = s.top();
        s.pop();
        ans += b * (long long)(b + 1) / 2 *
               (long long)(height[k] - max(height[i], height[s.top()]));
      }
      if (s.size() > 0 && height[s.top()] == height[i]) continue;
      height[k] = height[i];
      s.push(k);
    }
    ans += len - sa[len - 1] - height[len - 1];
    while (s.size() > 1) {
      int b = len - (s.top() - 1);
      k = s.top();
      s.pop();
      ans +=
          b * (long long)(b + 1) / 2 * (long long)(height[k] - height[s.top()]);
    }
    printf("%I64d\n", ans);
  }
  return 0;
}