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>
const int N = 100005, M = N * 2;
int tot, np, fa[M], len[M], ch[M][26], sz[M];
std::vector<int> g[M];
inline void extend(int x) {
  int p = np;
  np = ++tot, len[np] = len[p] + 1;
  memset(ch[np], 0, 104), sz[np] = 1;
  for (; p && !ch[p][x]; ch[p][x] = np, p = fa[p])
    ;
  if (!p)
    fa[np] = 1;
  else if (len[ch[p][x]] == len[p] + 1)
    fa[np] = ch[p][x];
  else {
    int nq = ++tot;
    len[nq] = len[p] + 1;
    sz[nq] = 0, memcpy(ch[nq], ch[ch[p][x]], 104);
    fa[nq] = fa[ch[p][x]], fa[np] = fa[ch[p][x]] = nq;
    for (int Q = ch[p][x]; p && ch[p][x] == Q; ch[p][x] = nq, p = fa[p])
      ;
  }
}
int n;
char s[N];
void solve() {
  scanf("%s", s + 1), n = strlen(s + 1);
  tot = np = 1;
  sz[1] = 0, memset(ch[1], 0, 104);
  for (int i = 1; i <= n; ++i) extend(s[i] - 'a');
  for (int i = 1; i <= tot; ++i) g[i].clear();
  for (int i = 2; i <= tot; ++i) g[fa[i]].emplace_back(i);
  long long ans = 0;
  std::function<void(int)> dfs = [&](int u) {
    for (auto v : g[u]) dfs(v), sz[u] += sz[v];
    ans += 1ll * sz[u] * sz[u] * (len[u] - len[fa[u]]);
  };
  dfs(1);
  printf("%lld\n", ans);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
  return 0;
}