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;
struct Node {
  int ch[26], len, fa;
  void clear() {
    for (int i = 0; i < 26; i++) ch[i] = 0;
    len = 0, fa = 0;
  }
} a[200055 << 1];
int lst, tot;
void add(int c) {
  int p = lst, np = ++tot;
  lst = np;
  a[np].len = a[p].len + 1;
  for (; p && !a[p].ch[c]; p = a[p].fa) a[p].ch[c] = np;
  if (!p)
    a[np].fa = 1;
  else {
    int v = a[p].ch[c];
    if (a[v].len == a[p].len + 1)
      a[np].fa = v;
    else {
      int nv = ++tot;
      a[nv] = a[v];
      a[nv].len = a[p].len + 1;
      for (; p && a[p].ch[c] == v; p = a[p].fa) a[p].ch[c] = nv;
      a[np].fa = a[v].fa = nv;
    }
  }
}
int size[200055], n, t;
vector<int> g[200055];
void dfs(int u) {
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    dfs(v);
    size[u] += size[v];
  }
}
char st[200055];
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%s", st + 1);
    n = strlen(st + 1);
    lst = tot = 1;
    for (int i = 1; i <= n; i++) add(st[i] - 'a');
    int p = 1;
    for (int i = 1; i <= n; i++) {
      p = a[p].ch[st[i] - 'a'];
      size[p]++;
    }
    for (int i = 2; i <= tot; i++) {
      g[a[i].fa].push_back(i);
    }
    dfs(1);
    long long ans = 0;
    for (int i = 1; i <= tot; i++)
      ans += 1ll * size[i] * size[i] * (a[i].len - a[a[i].fa].len);
    printf("%lld\n", ans);
    for (int i = 0; i <= tot; i++) g[i].clear(), size[i] = 0, a[i].clear();
  }
  return 0;
}