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 = 200005;
int n;
char s[maxn];
int tot, lst, cnt[maxn];
struct SAM_Node {
  int fail, len, ch[26];
} nd[maxn];
int newnode() {
  ++tot;
  nd[tot].fail = nd[tot].len = 0;
  for (int i = 0; i < 26; i++) nd[tot].ch[i] = 0;
  return tot;
}
void extend(int c) {
  int p = lst, np = newnode();
  cnt[lst = np] = 1;
  nd[np].len = nd[p].len + 1;
  for (; p && !nd[p].ch[c]; p = nd[p].fail) nd[p].ch[c] = np;
  if (!p)
    nd[np].fail = 1;
  else {
    int v = nd[p].ch[c];
    if (nd[p].len + 1 == nd[v].len)
      nd[np].fail = v;
    else {
      int nv = ++tot;
      nd[nv] = nd[v];
      nd[nv].len = nd[p].len + 1;
      for (; p && nd[p].ch[c] == v; p = nd[p].fail) nd[p].ch[c] = nv;
      nd[v].fail = nd[np].fail = nv;
    }
  }
}
int e, bg[maxn], to[maxn], nx[maxn];
void link(int u, int v) { to[++e] = v, nx[e] = bg[u], bg[u] = e; }
long long ans;
void dfs(int u) {
  for (int i = bg[u]; i; i = nx[i]) dfs(to[i]), cnt[u] += cnt[to[i]];
  if (u > 1 && cnt[u])
    ans += 1ll * cnt[u] * cnt[u] * (nd[u].len - nd[nd[u].fail].len);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    tot = lst = 0, tot = lst = newnode();
    memset(cnt, 0, sizeof(cnt));
    memset(bg, 0, sizeof(bg)), e = 0;
    scanf("%s", s), n = strlen(s);
    for (int i = 0; i < n; i++) extend(s[i] - 'a');
    for (int i = 2; i <= tot; i++) link(nd[i].fail, i);
    ans = 0, dfs(1);
    printf("%lld\n", ans);
  }
  return 0;
}