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 M = 1e5 + 5;
int T, n, tot, lst, id[M << 1], CB[M << 1], siz[M << 1];
char s[M];
struct Node {
  int f, len;
  int chi[26];
  Node() : f(0), len(0) { memset(chi, 0, 104); }
} SAM[M << 1];
inline void init() {
  while (tot) siz[tot] = 0, SAM[tot--] = Node();
  lst = tot = 1;
}
inline void Insert(const int& s) {
  int q, p, nq, np;
  p = lst;
  np = lst = ++tot;
  SAM[np].len = SAM[p].len + 1;
  siz[np] = 1;
  for (; p && !SAM[p].chi[s]; p = SAM[p].f) SAM[p].chi[s] = np;
  if (!p)
    SAM[np].f = 1;
  else {
    q = SAM[p].chi[s];
    if (SAM[p].len + 1 == SAM[q].len)
      SAM[np].f = q;
    else {
      SAM[nq = ++tot] = SAM[q];
      SAM[np].f = SAM[q].f = nq;
      SAM[nq].len = SAM[p].len + 1;
      for (; p && SAM[p].chi[s] == q; p = SAM[p].f) SAM[p].chi[s] = nq;
    }
  }
}
inline long long solve() {
  register long long ans = 0;
  register int i, now;
  for (i = 1; i <= tot; ++i) CB[i] = 0;
  for (i = 1; i <= tot; ++i) ++CB[SAM[i].len];
  for (i = 1; i <= tot; ++i) CB[i] += CB[i - 1];
  for (i = 1; i <= tot; ++i) id[CB[SAM[i].len]--] = i;
  for (i = tot; i >= 1; --i) {
    now = id[i];
    siz[SAM[now].f] += siz[now];
    ans += 1ll * (SAM[now].len - SAM[SAM[now].f].len) * siz[now] * siz[now];
  }
  return ans;
}
signed main() {
  scanf("%d", &T);
  while (T--) {
    init();
    scanf("%s", s);
    n = 0;
    while (isalpha(s[n])) Insert(s[n] - 97), s[n++] = 0;
    printf("%lld\n", solve());
  }
}