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 = 200010;
char a[maxn];
int wa[maxn], wb[maxn], wv[maxn], wts[maxn];
int cmp(int *r, int a, int b, int l) {
  return r[a] == r[b] && r[a + l] == r[b + l];
}
void init_sa(char *r, int *sa, int n, int m) {
  int i, j, *x = wa, *y = wb, *t;
  int p;
  for (i = 0; i < m; i++) wts[i] = 0;
  for (i = 0; i < n; i++) wts[x[i] = r[i]]++;
  for (i = 0; i < m; i++) wts[i] += wts[i - 1];
  for (i = n - 1; i >= 0; --i) sa[--wts[x[i]]] = i;
  for (j = 1, p = 1; p < n; j <<= 1, m = p) {
    for (p = 0, i = n - j; i < n; i++) y[p++] = i;
    for (i = 0; i < n; i++)
      if (sa[i] >= j) y[p++] = sa[i] - j;
    for (i = 0; i < n; i++) wv[i] = x[y[i]];
    for (i = 0; i < m; i++) wts[i] = 0;
    for (i = 0; i < n; i++) wts[wv[i]]++;
    for (i = 1; i < m; i++) wts[i] += wts[i - 1];
    for (i = n - 1; i >= 0; i--) sa[--wts[wv[i]]] = y[i];
    for (t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; i++)
      x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;
  }
  return;
}
int rnk[maxn], height[maxn];
int d[maxn][20], inx[maxn];
void init_inx() {
  inx[0] = -1;
  for (int i = 1; i < maxn; i++)
    inx[i] = (((i) & (i - 1)) ? 0 : 1) + inx[i - 1];
}
void init_rmq(int *r, int n) {
  for (int i = 0; i < n; i++) d[i][0] = height[i];
  for (int j = 1; j < 20; ++j) {
    int k = 1 << (j - 1);
    for (int i = 0; i + k * 2 - 1 < n; ++i)
      d[i][j] = min(d[i][j - 1], d[i + k][j - 1]);
  }
}
int query(int l, int r) {
  int k = inx[r - l + 1];
  return min(d[l][k], d[r - (1 << k) + 1][k]);
}
void calheight(char *r, int *sa, int n) {
  int i, j, k = 0;
  for (i = 1; i < n; i++) rnk[sa[i]] = i;
  for (i = 0; i < n - 1; height[rnk[i++]] = k)
    for (k ? k-- : 0, j = sa[rnk[i] - 1]; r[i + k] == r[j + k]; k++)
      ;
}
int sa[maxn];
int stk[maxn], tot = 0;
int lp[maxn], rp[maxn];
void solve() {
  scanf("%s", a);
  int len = strlen(a);
  a[len] = '$';
  len++;
  init_sa(a, sa, len, 256);
  calheight(a, sa, len);
  tot = 0;
  height[0] = height[len] = 0;
  for (int i = 1; i < len; i++) {
    while (tot > 0 && height[stk[tot - 1]] >= height[i]) tot--;
    lp[i] = tot == 0 ? 0 : stk[tot - 1];
    stk[tot++] = i;
  }
  tot = 0;
  for (int i = len - 1; i >= 1; i--) {
    while (tot > 0 && height[stk[tot - 1]] >= height[i]) tot--;
    rp[i] = tot == 0 ? len : stk[tot - 1];
    stk[tot++] = i;
  }
  long long ans = 0;
  priority_queue<pair<int, int>> que;
  for (int i = 1; i < len; i++) que.push({height[i], i});
  int prenum = -1, prehi = -1;
  for (int i = 1; i < len; i++)
    ans += 1ll * (len - 1 - sa[i]) - max(height[i], height[i + 1]);
  while (!que.empty()) {
    pair<int, int> u = que.top();
    que.pop();
    int hei = u.first, st = u.second;
    if (hei == prehi && lp[st] == lp[prenum]) continue;
    ans += 1ll * (rp[st] - lp[st] + 1) * (rp[st] - lp[st]) / 2 *
           (height[st] - max(height[lp[st]], height[rp[st]]));
    prehi = hei, prenum = st;
  }
  printf("%lld\n", ans);
}
int main(int argc, char const *argv[]) {
  solve();
  return 0;
}