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;
using ll = long long;
const int inf = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
struct SAM {
  int root, last, siz;
  int fa[maxn * 2], nxt[maxn * 2][26], len[maxn * 2], endpos[maxn * 2];
  int newnode(int _len) {
    fa[++siz] = 0;
    len[siz] = _len;
    endpos[siz] = 1;
    memset(nxt[siz], 0, sizeof nxt[siz]);
    return siz;
  }
  void init() {
    siz = 0;
    last = root = newnode(0);
  }
  void add(int x, int _len) {
    int p = last, np = newnode(_len);
    last = np;
    for (; p && !nxt[p][x]; p = fa[p]) nxt[p][x] = np;
    if (!p)
      fa[np] = root;
    else {
      int q = nxt[p][x];
      if (len[q] == len[p] + 1)
        fa[np] = q;
      else {
        int nq = ++siz;
        fa[nq] = fa[q];
        memcpy(nxt[nq], nxt[q], sizeof nxt[nq]);
        len[nq] = len[p] + 1;
        fa[q] = fa[np] = nq;
        for (; p && nxt[p][x] == q; p = fa[p]) nxt[p][x] = nq;
      }
    }
  }
  void build(string &s) {
    init();
    for (int i = 0; i < (int)s.size(); i++) add(s[i] - 'a', i + 1);
  }
  int v[maxn], toposam[maxn * 2];
  void toposort() {
    for (int i = 1; i <= siz; i++) v[len[i]]++;
    for (int i = 1; i <= len[last]; i++) v[i] += v[i - 1];
    for (int i = 1; i <= siz; i++) toposam[v[len[i]]--] = i;
  }
  void solve() {
    string s;
    cin >> s;
    build(s);
    toposort();
    ll ans = 0;
    for (int i = siz; i > 1; i--) {
      int u = toposam[i];
      endpos[fa[u]] += endpos[u];
      ans += (1LL * (len[u] - len[fa[u]]) * (endpos[u] + 1) * endpos[u]) >> 1;
    }
    cout << ans << "\n";
  }
} sam;
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  sam.solve();
  return 0;
}