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 iinf = 0x3f3f3f3f;
const long long linf = 0x3f3f3f3f3f3f3f3f;
const int N = 1e5;
int n;
string s;
const int tN = (N << 1) + 1;
int rt, t, tn, ch[tN][26], fa[tN], len[tN], sz[tN];
int newn() {
  fill(ch[tn], ch[tn] + 26, -1);
  fa[tn] = -1, sz[tn] = len[tn] = 0;
  return tn++;
}
void clear() { tn = 0, rt = t = newn(); }
void insert(int c) {
  int p = t, np = t = newn();
  len[np] = len[p] + 1;
  for (; ~p && !~ch[p][c]; p = fa[p]) ch[p][c] = np;
  if (!~p)
    fa[np] = rt;
  else {
    int q = ch[p][c];
    if (len[q] == len[p] + 1)
      fa[np] = q;
    else {
      int nq = newn();
      copy(ch[q], ch[q] + 26, ch[nq]);
      len[nq] = len[p] + 1, fa[nq] = fa[q], fa[q] = fa[np] = nq;
      for (; ~p && ch[p][c] == q; p = fa[p]) ch[p][c] = nq;
    }
  }
  sz[np] = 1;
}
int cnt[N + 1], q[tN];
void build() {
  fill(cnt, cnt + n + 1, 0);
  for (int i = (0), I = (tn); i < I; i++) cnt[len[i]]++;
  for (int i = (1), I = (n + 1); i < I; i++) cnt[i] += cnt[i - 1];
  for (int i = (0), I = (tn); i < I; i++) q[--cnt[len[i]]] = i;
  for (int i = (tn)-1, I = (0) - 1; i > I; i--)
    if (~fa[q[i]]) sz[fa[q[i]]] += sz[q[i]];
}
void Main() {
  cin >> s, n = int((s).size()), clear();
  for (int i = (0), I = (n); i < I; i++) insert(s[i] - 'a');
  build();
  long long ans = 0;
  for (int i = (1), I = (tn); i < I; i++)
    ans += 1ll * (len[i] - len[fa[i]]) * sz[i] * sz[i];
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int T;
  cin >> T;
  while (T--) Main();
  return 0;
}