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 = 3e5 + 5;
char s[maxn];
int mx[maxn];
int n;
struct SAM {
  int next[maxn][30], fa[maxn];
  long long len[maxn];
  int c[maxn], id[maxn];
  long long size[maxn];
  int root, tot, last;
  long long val(int x) { return len[x] - len[fa[x]]; }
  int newnode(int l) {
    fa[tot] = 0;
    for (int i = 0; i < 30; ++i) next[tot][i] = 0;
    len[tot++] = l;
    return tot - 1;
  }
  void init() {
    memset(len, 0, sizeof(len));
    memset(fa, 0, sizeof(fa));
    memset(size, 0, sizeof(size));
    memset(c, 0, sizeof(c));
    memset(id, 0, sizeof(id));
    tot = 1;
    last = root = newnode(0);
  }
  void extend(int x) {
    int p = last;
    int cur = newnode(len[p] + 1);
    size[cur] = 1;
    while (p != 0 && next[p][x] == 0) {
      next[p][x] = cur;
      p = fa[p];
    }
    if (p == 0)
      fa[cur] = root;
    else {
      int q = next[p][x];
      if (len[q] == len[p] + 1)
        fa[cur] = q;
      else {
        int tmp = newnode(len[p] + 1);
        memcpy(next[tmp], next[q], sizeof(next[q]));
        fa[tmp] = fa[q];
        fa[q] = fa[cur] = tmp;
        while (p != 0 && next[p][x] == q) {
          next[p][x] = tmp;
          p = fa[p];
        }
      }
    }
    last = cur;
  }
  void top_sort() {
    for (int i = 1; i <= tot; i++) ++c[len[i]];
    for (int i = 1; i <= tot; i++) c[i] += c[i - 1];
    for (int i = 1; i <= tot; i++) id[c[len[i]]--] = i;
  }
  void get_size() {
    for (int i = tot; i >= 1; i--) {
      int v = id[i];
      size[fa[v]] += size[v];
    }
  }
  long long solve() {
    top_sort();
    get_size();
    long long ans = 0;
    for (int i = 2; i <= tot; i++) {
      ans += 1ll * (size[i] * (size[i] + 1) / 2ll) * val(i);
    }
    return ans;
  }
} sam;
int main() {
  scanf("%s", s + 1);
  sam.init();
  int len = strlen(s + 1);
  for (int i = 1; i <= len; i++) {
    sam.extend(s[i] - 'a');
  }
  long long ans = sam.solve();
  printf("%lld\n", ans);
  return 0;
}