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;
template <class T>
inline void checkmin(T &a, const T &b) {
  if (b < a) a = b;
}
template <class T>
inline void checkmax(T &a, const T &b) {
  if (b > a) a = b;
}
const int maxn = 1e5 + 10;
int rk[maxn], sa[maxn], lcp[maxn], tmp[maxn], n, k;
string str;
stack<pair<int, int> > S;
bool compare_sa(int i, int j) {
  if (rk[i] != rk[j])
    return rk[i] < rk[j];
  else
    return (i + k < n ? rk[i + k] : -1) < (j + k < n ? rk[j + k] : -1);
}
void construct_sa(int *sa, string s) {
  n = (int)s.size();
  for (int i = 0; i <= n; i++) sa[i] = i, rk[i] = i < n ? s[i] : -1;
  for (k = 1; k <= n; k *= 2) {
    sort(sa, sa + n + 1, compare_sa);
    tmp[sa[0]] = 0;
    for (int i = 1; i <= n; i++)
      tmp[sa[i]] = tmp[sa[i - 1]] + (compare_sa(sa[i - 1], sa[i]) ? 1 : 0);
    for (int i = 0; i <= n; i++) rk[i] = tmp[i];
  }
}
void construct_lcp(int *sa, int *lcp, string s) {
  n = (int)s.size();
  for (int i = 0; i <= n; i++) rk[sa[i]] = i;
  int h = 0;
  lcp[0] = 0;
  for (int i = 0; i < n; i++) {
    int j = sa[rk[i] - 1];
    if (h > 0) h--;
    for (; i + h < n && j + h < n; h++)
      if (s[i + h] != s[j + h]) break;
    lcp[rk[i] - 1] = h;
  }
}
long long tree[maxn * 8], add[maxn * 8], st[maxn * 8];
void pushdown(int x, int l, int r) {
  int mid = (l + r) >> 1;
  if (st[x]) {
    tree[x << 1] = (mid - l + 1);
    tree[x << 1 | 1] = (r - mid);
    st[x << 1] = st[x << 1 | 1] = 1;
    add[x << 1] = add[x << 1 | 1] = 0;
    st[x] = 0;
  }
  if (add[x]) {
    tree[x << 1] += 1ll * (mid - l + 1) * add[x];
    tree[x << 1 | 1] += 1ll * (r - mid) * add[x];
    add[x << 1] += add[x];
    add[x << 1 | 1] += add[x];
    add[x] = 0;
  }
}
void updadd(int x, int l, int r, int cl, int cr, int val) {
  if (cr < l || r < cl) return;
  if (cl <= l && r <= cr) {
    pushdown(x, l, r);
    add[x] += val;
    tree[x] += 1ll * (r - l + 1) * val;
    return;
  }
  pushdown(x, l, r);
  int mid = (l + r) >> 1;
  updadd(x << 1, l, mid, cl, cr, val);
  updadd(x << 1 | 1, mid + 1, r, cl, cr, val);
  tree[x] = tree[x << 1] + tree[x << 1 | 1];
}
void updset(int x, int l, int r, int cl, int cr) {
  if (cr < l || r < cl) return;
  if (cl <= l && r <= cr) {
    pushdown(x, l, r);
    st[x] = 1;
    add[x] = 0;
    tree[x] = r - l + 1;
    return;
  }
  pushdown(x, l, r);
  int mid = (l + r) >> 1;
  updset(x << 1, l, mid, cl, cr);
  updset(x << 1 | 1, mid + 1, r, cl, cr);
  tree[x] = tree[x << 1] + tree[x << 1 | 1];
}
long long getval(int x, int l, int r, int cl, int cr) {
  if (cr < l || r < cl) return 0;
  if (cl <= l && r <= cr) return tree[x];
  pushdown(x, l, r);
  int mid = (l + r) >> 1;
  return getval(x << 1, l, mid, cl, cr) +
         getval(x << 1 | 1, mid + 1, r, cl, cr);
}
int main() {
  cin >> str;
  construct_sa(sa, str);
  construct_lcp(sa, lcp, str);
  long long ans = 0;
  for (int i = n; i >= 1; i--) {
    int len = lcp[i];
    updset(1, 0, n - 1, len, n - sa[i] - 1);
    if (len > 0) updadd(1, 0, n - 1, 0, len - 1, 1);
    ans += getval(1, 0, n - 1, 0, n - sa[i] - 1);
  }
  printf("%lld\n", ans);
}