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 N = 2e6 + 7;
int T, n, ch[N][26], fa[N], len[N], cnt[N];
int las, tot;
long long ans;
void copy(int x, int y) {
  for (int i = 0; i < 26; i++) ch[y][i] = ch[x][i];
  fa[y] = fa[x], len[y] = len[x];
}
void ins(int x) {
  int p = las, now = las = ++tot;
  len[now] = len[p] + 1, cnt[now] = 1;
  for (; p && !ch[p][x]; p = fa[p]) ch[p][x] = now;
  if (!p)
    fa[now] = 1;
  else {
    int pto = ch[p][x];
    if (len[pto] == len[p] + 1)
      fa[now] = pto;
    else {
      int sp = ++tot;
      copy(pto, sp), len[sp] = len[p] + 1;
      fa[now] = fa[pto] = sp;
      for (; p && ch[p][x] == pto; p = fa[p]) ch[p][x] = sp;
    }
  }
}
int c[N], q[N];
void work() {
  for (int i = 1; i <= tot; i++) c[len[i]]++;
  for (int i = 1; i <= tot; i++) c[i] += c[i - 1];
  for (int i = 1; i <= tot; i++) q[c[len[i]]--] = i;
  for (int i = tot; i >= 1; i--) cnt[fa[q[i]]] += cnt[q[i]];
  for (int i = 1; i <= tot; i++)
    ans += 1ll * cnt[i] * cnt[i] * (len[i] - len[fa[i]]);
}
char s[N];
int main() {
  scanf("%d", &T);
  while (T--) {
    ans = 0;
    memset(c, 0, sizeof(c));
    memset(q, 0, sizeof(q));
    memset(ch, 0, sizeof(ch));
    memset(len, 0, sizeof(len));
    memset(cnt, 0, sizeof(cnt));
    memset(fa, 0, sizeof(fa));
    las = 1, tot = 1;
    scanf("%s", s);
    n = strlen(s);
    for (int i = 0; i < n; i++) ins(s[i] - 'a');
    work();
    printf("%lld\n", ans);
  }
  return 0;
}