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.
#include <bits/stdc++.h>
using namespace std;
const int N = 200500;
struct state {
  int len, link;
  map<char, int> next;
};
state st[N];
int sz, last;
void sa_init() {
  last = st[0].len = 0;
  sz = 1;
  st[0].link = -1;
}
void sa_extend(char c) {
  int k = sz++, p;
  st[k].len = st[last].len + 1;
  for (p = last; p != -1 && !st[p].next.count(c); p = st[p].link)
    st[p].next[c] = k;
  if (p == -1)
    st[k].link = 0;
  else {
    int q = st[p].next[c];
    if (st[p].len + 1 == st[q].len)
      st[k].link = q;
    else {
      int w = sz++;
      st[w].len = st[p].len + 1;
      st[w].next = st[q].next;
      st[w].link = st[q].link;
      for (; p != -1 && st[p].next[c] == q; p = st[p].link) st[p].next[c] = w;
      st[q].link = st[k].link = w;
    }
  }
  last = k;
}
string s;
long long dp[N];
bool t[N];
long long f(int i) {
  if (dp[i] != -1) return dp[i];
  long long &r = dp[i] = t[i];
  for (auto p : st[i].next) r += f(p.second);
  return r;
}
void solve() {
  sa_init();
  memset(dp, -1, sizeof dp);
  for (int i = 0, qwerty = ((int)(s).size()); i < qwerty; i++) sa_extend(s[i]);
  int x = last;
  while (x) {
    t[x] = 1;
    x = st[x].link;
  }
  f(0);
  long long r = 0;
  for (int i = 1, qwerty = sz; i < qwerty; i++)
    r += f(i) * f(i) * (st[i].len - st[st[i].link].len);
  while (last) {
    t[last] = 0;
    last = st[last].link;
  }
  for (int i = 0, qwerty = sz; i < qwerty; i++) st[i].next.clear();
  cout << r << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tn;
  cin >> tn;
  for (int i = 0, qwerty = tn; i < qwerty; i++) {
    cin >> s;
    solve();
  }
}
