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 = 100005;
const int inf = 0x3f3f3f3f;
namespace suffix_array {
int n, k;
int rank[maxn], tmp[maxn];
int level[maxn];
bool compare_sa(int i, int j) {
  if (rank[i] != rank[j])
    return rank[i] < rank[j];
  else {
    int ri = i + k <= n ? rank[i + k] : -1;
    int rj = j + k <= n ? rank[j + k] : -1;
    return ri < rj;
  }
}
void construct_sa(char* s, int* sa) {
  n = strlen(s);
  for (int i = 0; i <= n; i++) {
    sa[i] = i;
    rank[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++) {
      rank[i] = tmp[i];
    }
  }
}
void construct_lcp(char* s, int* sa, int* lcp) {
  for (int i = 0; i <= n; i++) level[sa[i]] = i;
  int h = 0;
  lcp[0] = 0;
  for (int i = 0; i < n; i++) {
    int j = sa[rank[i] - 1];
    if (h > 0) h--;
    for (; j + h < n && i + h < n; h++) {
      if (s[j + h] != s[i + h]) break;
    }
    lcp[rank[i] - 1] = h;
  }
}
}  // namespace suffix_array
int sa[maxn], lcp[maxn];
int hight[maxn];
int dp[maxn][30];
char str[maxn];
void rmq_init(int n) {
  for (int i = 1; i <= n; i++) dp[i][0] = hight[i];
  for (int j = 1; (1 << j) <= n; j++)
    for (int i = 1; i + (1 << j) - 1 <= n; i++)
      dp[i][j] = min(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
}
int rmq(int l, int r) {
  if (l > r) swap(l, r);
  int k = 0;
  while (1 << (k + 1) <= r - l + 1) k++;
  int res = min(dp[l][k], dp[r - (1 << k) + 1][k]);
  return res;
}
void solve(char* s) {
  long long n = strlen(s);
  long long ans = n * (n + 1) / 2;
  suffix_array::construct_sa(s, sa);
  suffix_array::construct_lcp(s, sa, lcp);
  for (int i = 0; i < n; i++) hight[i + 1] = lcp[i];
  rmq_init(n);
  for (int i = 1; i <= n; i++) {
    int l = 0, r = i;
    while (r > l + 1) {
      int mid = (l + r) / 2;
      if (rmq(mid, i - 1) >= hight[i])
        r = mid;
      else
        l = mid;
    }
    int x = i, y = n + 1;
    while (y > x + 1) {
      int mid = (x + y) / 2;
      if (rmq(i + 1, mid) > hight[i])
        x = mid;
      else
        y = mid;
    }
    ans += 1LL * hight[i] * (x - i + 1) * (i - r + 1);
  }
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> str;
  solve(str);
  return 0;
}