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 = 1e5 + 5;
char s[maxn];
int n;
struct SAM {
  int next[maxn << 1][26], link[maxn << 1], len[maxn << 1];
  int a[maxn << 1], id[maxn << 1], right[maxn << 1];
  int tot, root, last, num[maxn << 1];
  inline void init() {
    for (int i = 0; i < tot + 1; ++i)
      memset(next[i], 0, sizeof(next[i])), right[i] = 0;
    memset(len, 0, sizeof(len));
    memset(link, 0, sizeof(link));
    tot = last = root = 1;
  }
  inline void extend(int x) {
    int p = last, now = ++tot;
    right[now] = 1;
    len[now] = len[p] + 1;
    while (p && !next[p][x]) next[p][x] = now, p = link[p];
    if (!p)
      link[now] = root;
    else {
      int q = next[p][x];
      if (len[q] == len[p] + 1)
        link[now] = q;
      else {
        int tmp = ++tot;
        memcpy(next[tmp], next[q], sizeof(next[q]));
        link[tmp] = link[q];
        link[q] = link[now] = tmp;
        len[tmp] = len[p] + 1;
        while (p && next[p][x] == q) next[p][x] = tmp, p = link[p];
      }
    }
    last = now;
  }
  inline void solve() {
    for (int i = 1; i < tot + 1; ++i) ++a[len[i]];
    for (int i = 1; i < tot + 1; ++i) a[i] += a[i - 1];
    for (int i = 1; i < tot + 1; ++i) id[a[len[i]]--] = i;
    int last = 1;
    for (int i = 0; i < n; ++i) {
      num[next[last][s[i] - 'a']] = 1;
      last = next[last][s[i] - 'a'];
    }
    long long ans(0);
    for (int i = tot; i >= 1; --i) {
      num[link[id[i]]] += num[id[i]];
      ans += 1ll * (len[id[i]] - len[link[id[i]]]) * (num[id[i]] + 1) *
             num[id[i]] / 2;
    }
    printf("%lld\n", ans);
  }
} sam;
void solve() {
  scanf("%s", s);
  n = strlen(s);
  sam.init();
  for (int i = 0; i < n; ++i) sam.extend(s[i] - 'a');
  sam.solve();
}
int main() { solve(); }