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 = 300005;
int tot = 1, last = 1, link[MAXN], ch[MAXN][26], len[MAXN], siz[MAXN];
long long ans = 0;
char s[MAXN];
vector<int> G[MAXN];
inline void extend(int c) {
  int now = ++tot;
  len[now] = len[last] + 1, siz[now] = 1;
  int p = last;
  last = now;
  while (p && !ch[p][c]) ch[p][c] = now, p = link[p];
  if (!p) {
    link[now] = 1;
    return;
  }
  int q = ch[p][c];
  if (len[q] == len[p] + 1) {
    link[now] = q;
    return;
  }
  int newq = ++tot;
  len[newq] = len[p] + 1;
  memcpy(ch[newq], ch[q], sizeof(ch[q]));
  link[newq] = link[q];
  link[q] = link[now] = newq;
  while (p && ch[p][c] == q) ch[p][c] = newq, p = link[p];
}
inline void cl() {
  tot = 1, last = 1, ans = 0;
  memset(link, 0, sizeof(link));
  memset(ch, 0, sizeof(ch));
  memset(len, 0, sizeof(len));
  memset(siz, 0, sizeof(siz));
  for (int i = 0; i < MAXN; i++) G[i].clear();
}
void dfs(int u) {
  for (int v : G[u]) dfs(v), siz[u] += siz[v];
  ans += 1ll * siz[u] * siz[u] * (len[u] - len[link[u]]);
}
inline void solve() {
  cl();
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  for (int i = 1; i <= n; i++) extend(s[i] - 'a');
  for (int i = 2; i <= tot; i++) G[link[i]].push_back(i);
  dfs(1);
  printf("%lld\n", ans);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}