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;
long long first[500005], nxt[500005], to[500005], tot = 0;
void build(long long x, long long y) {
  nxt[++tot] = first[x];
  first[x] = tot;
  to[tot] = y;
}
struct SAM {
  long long ch[26], len, link;
} st[500005];
long long Len, lst, sz, ans, f[500005];
char ch[500005];
void Add(long long c) {
  long long p = lst, cur = ++sz;
  lst = cur;
  f[lst] = 1;
  st[cur].len = st[p].len + 1;
  while (p != -1 && !st[p].ch[c]) st[p].ch[c] = cur, p = st[p].link;
  if (p == -1) return;
  long long q = st[p].ch[c];
  if (st[p].len + 1 == st[q].len) {
    st[cur].link = q;
    return;
  }
  long long clone = ++sz;
  st[clone] = st[q];
  st[clone].len = st[p].len + 1;
  while (p != -1 && st[p].ch[c] == q) st[p].ch[c] = clone, p = st[p].link;
  st[q].link = st[cur].link = clone;
}
void dfs(long long u) {
  for (long long e = first[u]; e; e = nxt[e]) dfs(to[e]), f[u] += f[to[e]];
  if (!u) return;
  ans += (st[u].len - st[st[u].link].len) * f[u] * f[u];
}
signed main() {
  long long T;
  cin >> T;
  while (T--) {
    memset(st, 0, sizeof(st));
    memset(first, 0, sizeof(first));
    memset(f, 0, sizeof(f));
    sz = lst = ans = tot = 0;
    st[0].link = -1;
    scanf("%s", ch + 1);
    Len = strlen(ch + 1);
    for (long long i = 1; i <= Len; i++) Add(ch[i] - 'a');
    for (long long i = 1; i <= sz; i++) build(st[i].link, i);
    dfs(0);
    cout << ans << endl;
  }
}