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 N = 200000 + 10;
const int INF = 1e9 + 7;
int cc[N], sa[N], t1[N], t2[N], height[N], rk[N];
int n;
char s[N];
bool cmp(int *y, int a, int b, int k) {
  int a1 = y[a], a2 = a + k >= n ? -1 : y[a + k];
  int b1 = y[b], b2 = b + k >= n ? -1 : y[b + k];
  return a1 == b1 && a2 == b2;
}
void buildSa(int m) {
  int *x = t1, *y = t2;
  for (int i = 0; i < m; i++) cc[i] = 0;
  for (int i = 0; i < n; i++) cc[x[i] = s[i]]++;
  for (int i = 1; i < m; i++) cc[i] += cc[i - 1];
  for (int i = n - 1; i >= 0; i--) sa[--cc[x[i]]] = i;
  for (int k = 1; k <= n; k <<= 1) {
    int p = 0;
    for (int i = n - k; i < n; i++) y[p++] = i;
    for (int i = 0; i < n; i++)
      if (sa[i] >= k) y[p++] = sa[i] - k;
    for (int i = 0; i < m; i++) cc[i] = 0;
    for (int i = 0; i < n; i++) cc[x[y[i]]]++;
    for (int i = 1; i < m; i++) cc[i] += cc[i - 1];
    for (int i = n - 1; i >= 0; i--) sa[--cc[x[y[i]]]] = y[i];
    swap(x, y);
    m = 1, x[sa[0]] = 0;
    for (int i = 1; i < n; i++) {
      x[sa[i]] = cmp(y, sa[i - 1], sa[i], k) ? m - 1 : m++;
    }
    if (m >= n) break;
  }
}
void buildH() {
  for (int i = 0; i < n; i++) rk[sa[i]] = i;
  int k = 0;
  height[0] = 0;
  for (int i = 0; i < n; i++) {
    if (!rk[i]) continue;
    if (k) k--;
    int j = sa[rk[i] - 1];
    while (s[i + k] == s[j + k]) k++;
    height[rk[i]] = k;
  }
}
pair<int, int> stk[N];
int top;
int pre[N], nex[N];
int main() {
  scanf("%s", s);
  n = strlen(s);
  buildSa(1000);
  buildH();
  top = 0, stk[++top] = make_pair(0, -INF);
  for (int i = 1; i < n; i++) {
    while (stk[top].second >= height[i]) --top;
    pre[i] = stk[top].first;
    stk[++top] = make_pair(i, height[i]);
  }
  top = 0, stk[++top] = make_pair(n, -INF);
  for (int i = n - 1; i >= 0; i--) {
    while (stk[top].second > height[i]) --top;
    nex[i] = stk[top].first;
    stk[++top] = make_pair(i, height[i]);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += n - sa[i];
  }
  for (int i = 1; i < n; i++) {
    ans += 1LL * (i - pre[i]) * (nex[i] - i) * height[i];
  }
  printf("%lld\n", ans);
}