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 fw[200005];
int fwlen, lst, tot, head[200005], sze;
struct node {
  int to, nxt;
} e[200005 * 2];
struct node2 {
  int nxt[27], len, link;
} st[200005 * 2];
void build(int a, int b) {
  e[++tot].to = b;
  e[tot].nxt = head[a];
  head[a] = tot;
}
long long ans, f[200005 * 2];
void add(int c) {
  int p = lst, cur = ++sze;
  f[cur] = 1;
  lst = cur;
  st[cur].len = st[p].len + 1;
  while (p != -1 && !st[p].nxt[c]) st[p].nxt[c] = cur, p = st[p].link;
  if (p == -1) {
    return;
  }
  int q = st[p].nxt[c];
  if (st[p].len + 1 == st[q].len) {
    st[cur].link = q;
    return;
  }
  int clone = ++sze;
  st[clone] = st[q];
  st[clone].len = st[p].len + 1;
  while (p != -1 && st[p].nxt[c] == q) {
    st[p].nxt[c] = clone;
    p = st[p].link;
  }
  st[q].link = st[cur].link = clone;
}
void dfs(int p) {
  for (int i = head[p]; i; i = e[i].nxt) {
    dfs(e[i].to);
    f[p] += f[e[i].to];
  }
  if (p == 0) return;
  ans += f[p] * f[p] * (st[p].len - st[st[p].link].len);
}
int main() {
  int n;
  scanf("%d", &n);
  while (n--) {
    tot = sze = ans = lst = 0;
    memset(f, 0, sizeof f);
    memset(st, 0, sizeof st);
    memset(head, 0, sizeof head);
    scanf("%s", fw);
    fwlen = strlen(fw);
    st[0].link = -1;
    for (int i = 0; i < fwlen; ++i) {
      add(fw[i] - 'a');
    }
    for (int i = 1; i <= sze; ++i) build(st[i].link, i);
    dfs(0);
    printf("%lld\n", ans);
  }
  return 0;
}