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;
inline long long read() {
  long long x = 0;
  int f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + (ch ^ 48);
  return x * f;
}
void print(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
const int N = 100010, SIZ = N * 2;
struct SAM {
  struct state {
    int len, link, nxt[26];
  } s[SIZ];
  int cnt, lst;
  long long siz[SIZ], ans;
  struct edge {
    int to, nxt;
  } e[SIZ];
  int head[SIZ], tot;
  void add(int u, int v) {
    e[++tot].to = v, e[tot].nxt = head[u];
    head[u] = tot;
  }
  SAM() { lst = cnt = 1; }
  void insert(int c) {
    int cur = ++cnt;
    siz[cur] = 1;
    s[cur].len = s[lst].len + 1;
    int p = lst;
    for (; p && !s[p].nxt[c]; p = s[p].link) s[p].nxt[c] = cur;
    if (!p)
      s[cur].link = 1;
    else {
      int q = s[p].nxt[c];
      if (s[q].len == s[p].len + 1)
        s[cur].link = q;
      else {
        int clone = ++cnt;
        s[clone] = s[q];
        s[clone].len = s[p].len + 1;
        for (; p && s[p].nxt[c] == q; p = s[p].link) s[p].nxt[c] = clone;
        s[q].link = s[cur].link = clone;
      }
    }
    lst = cur;
  }
  void build() {
    for (int i = 2; i <= cnt; i++) add(s[i].link, i);
    ans = 0;
  }
  void dfs(int u) {
    for (int i = head[u]; i; i = e[i].nxt) dfs(e[i].to), siz[u] += siz[e[i].to];
    ans += siz[u] * (siz[u] + 1) / 2 * (s[u].len - s[s[u].link].len);
  }
} S;
char s[N];
int main() {
  scanf("%s", s + 1);
  for (int i = 1; s[i]; i++) S.insert(s[i] - 'a');
  S.build();
  S.dfs(1);
  ::print(S.ans), putchar('\n');
  return 0;
}