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;
int t, n, m, cur;
int nxt[200200][26], link[200200], dp[200200], cnt[200200], ord[200200],
    occur[200200];
char s[200200];
void extend(int c) {
  int x = cur, y = cur = ++m;
  dp[y] = dp[x] + 1;
  occur[y] = 1;
  while (x && !nxt[x][c]) nxt[x][c] = y, x = link[x];
  if (!x)
    link[y] = 1;
  else {
    int z = nxt[x][c];
    if (dp[z] == dp[x] + 1)
      link[y] = z;
    else {
      int u = ++m;
      memcpy(nxt[u], nxt[z], sizeof nxt[u]);
      link[u] = link[z], link[z] = link[y] = u;
      dp[u] = dp[x] + 1;
      while (x && nxt[x][c] == z) nxt[x][c] = u, x = link[x];
    }
  }
}
void build(char *s) {
  memset(nxt, 0, sizeof nxt), memset(link, 0, sizeof link),
      memset(dp, 0, sizeof dp);
  n = strlen(s), m = 0;
  cur = ++m;
  for (int i = 0; i < n; i++) extend(s[i] - 'a');
  memset(cnt, 0, sizeof cnt);
  for (int i = 1; i <= m; i++) cnt[dp[i]]++;
  for (int i = 1; i <= n; i++) cnt[i] += cnt[i - 1];
  for (int i = m; i >= 1; i--) ord[cnt[dp[i]]--] = i;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%s", s);
    memset(occur, 0, sizeof occur);
    build(s);
    for (int i = m; i >= 1; i--) occur[link[ord[i]]] += occur[ord[i]];
    long long ans = 0;
    for (int i = 1; i <= m; i++)
      ans += 1ll * occur[i] * occur[i] * (dp[i] - dp[link[i]]);
    printf("%I64d\n", ans);
  }
  return 0;
}