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 INF = 0x3f3f3f3f;
const double eps = 1e-8;
const double PI = acos(-1.0);
const int MAXN = 1e5 + 5;
const int MAXM = 2e6 + 5;
const int MOD = 1e9 + 7;
int sgn(double x) {
  if (fabs(x) < eps) return 0;
  if (x < 0)
    return -1;
  else
    return 1;
}
long long pw(long long a, long long n, long long mod) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = ret * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return ret;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
void inc(long long &a, long long b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
void dec(long long &a, long long b) {
  a -= b;
  if (a < 0) a += MOD;
}
int mul(int a, int b) {
  long long c = 1ll * a * b;
  return c - c / MOD * MOD;
}
struct SAM {
  int fail[MAXN * 2], nxt[MAXN * 2][26];
  int len[MAXN * 2], right[MAXN * 2];
  int cnt;
  int new_node(int v) {
    ++cnt;
    fail[cnt] = 0, right[cnt] = 0, len[cnt] = v;
    for (int i = 0; i < 26; i++) nxt[cnt][i] = 0;
    return cnt;
  }
  int root, last;
  SAM() { clear(); }
  void clear() { root = last = new_node(cnt = 0); }
  void add(int c) {
    int p = last, np = new_node(len[p] + 1);
    while (p && !nxt[p][c]) nxt[p][c] = np, p = fail[p];
    if (!p)
      fail[np] = root;
    else {
      int q = nxt[p][c];
      if (len[p] + 1 == len[q])
        fail[np] = q;
      else {
        int nq = new_node(len[p] + 1);
        for (int i = 0; i < 26; i++) nxt[nq][i] = nxt[q][i];
        fail[nq] = fail[q], fail[q] = fail[np] = nq;
        while (p && nxt[p][c] == q) nxt[p][c] = nq, p = fail[p];
      }
    }
    last = np, right[np] = 1;
  }
  int d[MAXN * 2], b[MAXN * 2];
  void topo() {
    for (int i = 0; i < cnt + 1; i++) d[i] = 0;
    int mx = 0;
    for (int i = 1; i < cnt + 1; i++) mx = max(mx, len[i]), d[len[i]]++;
    for (int i = 1; i < mx + 1; i++) d[i] += d[i - 1];
    for (int i = 1; i < cnt + 1; i++) b[d[len[i]]--] = i;
  }
  void upd() {
    topo();
    for (int i = cnt + 1 - 1; i >= 2; i--) right[fail[b[i]]] += right[b[i]];
  }
  void work() {
    long long ans = 0;
    for (int i = 2; i < cnt + 1; i++)
      ans += 1ll * (len[i] - len[fail[i]]) * (right[i] + 1) * right[i] / 2;
    printf("%lld\n", ans);
  }
} sm;
char s[MAXN];
int main() {
  scanf("%s", s);
  int n = strlen(s);
  for (int i = 0; i < n; i++) sm.add(s[i] - 'a');
  sm.upd();
  sm.work();
}