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 = 2e5 + 5;
char s[MAXN];
int lst = 1, cnt = 1, len[MAXN];
int fa[MAXN], ch[MAXN][26];
int siz[MAXN], c[MAXN], ord[MAXN];
inline int read() {
  int x = 0;
  char ch = getchar();
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x;
}
inline void insert(int c) {
  int p = lst, np = lst = ++cnt;
  len[np] = len[p] + 1, siz[np] = 1;
  while (p && !ch[p][c]) ch[p][c] = np, p = fa[p];
  if (!p) return fa[np] = 1, void();
  int q = ch[p][c];
  if (len[q] == len[p] + 1) return fa[np] = q, void();
  int nq = ++cnt;
  fa[nq] = fa[q], fa[q] = fa[np] = nq;
  len[nq] = len[p] + 1, memcpy(ch[nq], ch[q], sizeof(ch[q]));
  while (ch[p][c] == q) ch[p][c] = nq, p = fa[p];
}
inline void clear() {
  lst = cnt = 1;
  memset(c, 0, sizeof(c));
  memset(fa, 0, sizeof(fa));
  memset(ch, 0, sizeof(ch));
  memset(siz, 0, sizeof(siz));
}
inline long long solve() {
  long long ans = 0;
  for (int i = 1; i <= cnt; i++) c[len[i]]++;
  for (int i = 1; i <= cnt; i++) c[i] += c[i - 1];
  for (int i = cnt; i; i--) ord[c[len[i]]--] = i;
  for (int i = cnt; i; i--) siz[fa[ord[i]]] += siz[ord[i]];
  for (int i = 1; i <= cnt; i++)
    ans += 1ll * (len[i] - len[fa[i]]) * siz[i] * siz[i];
  return ans;
}
int main() {
  int t = read();
  while (t--) {
    clear();
    scanf("%s", s + 1);
    int n = strlen(s + 1);
    for (int i = 1; i <= n; i++) insert(s[i] - 'a');
    printf("%lld\n", solve());
  }
  return 0;
}