Problem: Now that you have proposed a fake post for the HC2 Facebook page, Heidi wants to measure the quality of the post before actually posting it. She recently came across a (possibly fake) article about the impact of fractal structure on multimedia messages and she is now trying to measure the self-similarity of the message, which is defined as

<image>

where the sum is over all nonempty strings p and <image> is the number of occurences of p in s as a substring. (Note that the sum is infinite, but it only has a finite number of nonzero summands.)

Heidi refuses to do anything else until she knows how to calculate this self-similarity. Could you please help her? (If you would like to instead convince Heidi that a finite string cannot be a fractal anyway – do not bother, we have already tried.)

Input

The input starts with a line indicating the number of test cases T (1 ≤ T ≤ 10). After that, T test cases follow, each of which consists of one line containing a string s (1 ≤ |s| ≤ 100 000) composed of lowercase letters (a-z).

Output

Output T lines, every line containing one number – the answer to the corresponding test case.

Example

Input

4
aa
abcd
ccc
abcc


Output

5
10
14
12

Note

A string s contains another string p as a substring if p is a contiguous subsequence of s. For example, ab is a substring of cab but not of acb.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
const int LG = 20;
const int MOD = 1e9 + 7;
string s;
pair<pair<int, int>, int> arr[MAXN];
int sufrank[LG][MAXN], n;
int revrank[MAXN], lcp[MAXN];
int L[MAXN], R[MAXN];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int TT;
  cin >> TT;
  while (TT--) {
    memset(sufrank, 0, sizeof(sufrank));
    memset(revrank, 0, sizeof(revrank));
    memset(lcp, 0, sizeof(lcp));
    cin >> s;
    n = s.size();
    revrank[n] = -1;
    for (int i = 0; i < n; i++) sufrank[0][i] = s[i] - 'a' + 256;
    for (int l = 1; l < LG; l++) {
      for (int i = 0, j = 1 << (l - 1); i < n; i++, j++)
        arr[i] = {{sufrank[l - 1][i], j < n ? sufrank[l - 1][j] : -256}, i};
      sort(arr, arr + n);
      for (int i = 1; i < n; i++)
        sufrank[l][arr[i].second] = (arr[i].first == arr[i - 1].first)
                                        ? sufrank[l][arr[i - 1].second]
                                        : sufrank[l][arr[i - 1].second] + 1;
    }
    for (int i = 0; i < n; i++) revrank[sufrank[LG - 1][i]] = i;
    for (int i = 0, E = 0; i < n; i++) {
      int A = revrank[sufrank[LG - 1][i] + 1];
      if (A == -1) {
        E = 0;
        continue;
      }
      while ((A + E) < n && (i + E) < n && s[A + E] == (s[i + E])) E++;
      lcp[sufrank[LG - 1][i]] = E;
      if (E) E--;
    }
    vector<int> vec;
    for (int i = 0; i < (n - 1); i++) {
      while (vec.size() && lcp[vec.back()] > lcp[i]) vec.pop_back();
      L[i] = vec.size() ? vec.back() : -1;
      vec.push_back(i);
    }
    vec.clear();
    for (int i = n - 2; ~i; i--) {
      while (vec.size() && lcp[vec.back()] >= lcp[i]) vec.pop_back();
      R[i] = vec.size() ? vec.back() : n - 1;
      vec.push_back(i);
    }
    long long answer = 1ll * n * (n + 1) / 2;
    for (int i = 0; i < (n - 1); i++)
      answer += lcp[i] * (2ll * (i - L[i]) * (R[i] - i));
    cout << answer << endl;
  }
  return 0;
}