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 N = 1e5;
struct suffixAutomata {
  struct node {
    int len, link, next[26];
  } aut[N + N + 2];
  int sz, last, occur[N + N + 2];
  vector<int> tree[N + N + 2];
  void clean(int n = N) {
    for (int i = 0; i <= n + n; i++) {
      memset(aut[i].next, 0, sizeof(aut[i].next));
      tree[i].clear();
      occur[i] = 0;
    }
    sz = 0, last = 0, aut[0].link = -1;
  }
  void insrt(int c) {
    int now = ++sz;
    aut[now].len = aut[last].len + 1;
    int p, q, cl;
    occur[now] = 1;
    for (p = last; p != -1 && !aut[p].next[c]; p = aut[p].link)
      aut[p].next[c] = now;
    if (p == -1)
      aut[now].link = 0;
    else {
      q = aut[p].next[c];
      if (aut[p].len + 1 == aut[q].len)
        aut[now].link = q;
      else {
        cl = ++sz;
        aut[cl].len = aut[p].len + 1;
        aut[cl].link = aut[q].link;
        memcpy(aut[cl].next, aut[q].next, sizeof(aut[cl].next));
        for (; p != -1 && aut[p].next[c] == q; p = aut[p].link)
          aut[p].next[c] = cl;
        aut[now].link = aut[q].link = cl;
      }
    }
    last = now;
  }
  void dfs(int now) {
    for (auto x : tree[now]) {
      dfs(x);
      occur[now] += occur[x];
    }
  }
  void makeTree(void) {
    for (int i = 1; i <= sz; i++) tree[aut[i].link].push_back(i);
    dfs(0);
  }
  long long cal(void) {
    long long ans = 0;
    for (int i = 1; i <= sz; i++) {
      long long sub = aut[i].len - aut[aut[i].link].len;
      ans += sub * occur[i] * occur[i];
    }
    return ans;
  }
} sam;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    string s;
    cin >> s;
    sam.clean(s.size());
    for (auto x : s) sam.insrt(x - 'a');
    sam.makeTree();
    cout << sam.cal() << "\n";
  }
  return 0;
}