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;
char s[200010];
int tmp[200010], len[200010], fa[200010], ch[200010][26], sz[200010], last,
    node, id[200010];
void init() {
  last = node = 1;
  memset(ch, 0, sizeof(ch));
  memset(sz, 0, sizeof(sz));
  memset(tmp, 0, sizeof(tmp));
  memset(fa, 0, sizeof(fa));
}
void insert(int c) {
  int f = last, p = ++node;
  last = p;
  len[p] = len[f] + 1;
  sz[p] = 1;
  while (f && !ch[f][c]) ch[f][c] = p, f = fa[f];
  if (!f)
    fa[p] = 1;
  else {
    int x = ch[f][c];
    if (len[f] + 1 == len[x])
      fa[p] = x;
    else {
      int nx = ++node;
      len[nx] = len[f] + 1;
      fa[nx] = fa[x];
      memcpy(ch[nx], ch[x], sizeof(ch[nx]));
      fa[x] = fa[p] = nx;
      while (f && ch[f][c] == x) ch[f][c] = nx, f = fa[f];
    }
  }
}
int main() {
  int cs;
  cin >> cs;
  while (cs--) {
    init();
    scanf("%s", s);
    int l = strlen(s);
    for (int i = 0; i < l; i++) insert(s[i] - 'a');
    for (int i = 1; i <= node; i++) tmp[len[i]]++;
    for (int i = 1; i <= node; i++) tmp[i] += tmp[i - 1];
    for (int i = 1; i <= node; i++) id[tmp[len[i]]--] = i;
    long long ans = 0;
    for (int i = node; i >= 1; i--) {
      int now = id[i];
      sz[fa[now]] += sz[now];
      ans += 1ll * sz[now] * sz[now] * (len[now] - len[fa[now]]);
    }
    cout << ans << "\n";
  }
  return 0;
}