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 s[100005];
struct SAM {
  int len;
  int last;
  int son[27];
} t[2000005];
struct edge {
  int obj;
  int last;
} e[400050];
int eg;
int head[200050];
void link(int u, int v) {
  e[++eg].obj = v;
  e[eg].last = head[u];
  head[u] = eg;
}
int g, u;
int cnt[200005];
void Insert(char c) {
  int v = ++g;
  t[v].last = 0;
  t[v].len = t[u].len + 1;
  cnt[v] = 1;
  for (int i = 0; i < 26; i++) t[v].son[i] = 0;
  int p = u;
  while (p != -1 && !t[p].son[c - 'a']) {
    t[p].son[c - 'a'] = v;
    p = t[p].last;
  }
  if (p == -1)
    t[v].last = 0;
  else {
    int q = t[p].son[c - 'a'];
    if (t[p].len + 1 == t[q].len) {
      t[v].last = q;
    } else {
      int copy = ++g;
      t[copy].len = t[p].len + 1;
      for (int i = 0; i < 26; i++) t[copy].son[i] = t[q].son[i];
      t[copy].last = t[q].last;
      while (p != -1 && t[p].son[c - 'a'] == q) {
        t[p].son[c - 'a'] = copy;
        p = t[p].last;
      }
      t[q].last = t[v].last = copy;
    }
  }
  u = v;
}
long long ans = 0;
void dfs(int now) {
  for (int i = head[now]; i; i = e[i].last) {
    int v = e[i].obj;
    dfs(v);
    cnt[now] += cnt[v];
  }
  if (now)
    ans += (long long)(cnt[now]) * (long long)(cnt[now]) *
           (long long)(t[now].len - t[t[now].last].len);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    g = 0;
    u = 0;
    t[u].len = 0, t[u].last = -1;
    cnt[0] = 1;
    for (int i = 0; i < 26; i++) t[u].son[i] = 0;
    memset(cnt, 0, sizeof(cnt));
    memset(head, 0, sizeof(head));
    eg = 0;
    ans = 0;
    scanf("%s", s + 1);
    int kkk = strlen(s + 1);
    for (int i = 1; i <= kkk; i++) {
      Insert(s[i]);
    }
    for (int i = 1; i <= g; i++) link(t[i].last, i);
    dfs(0);
    printf("%lld\n", ans);
  }
  return 0;
}