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;
template <class T>
void read(T& x) {
  T res = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    res = (res << 3) + (res << 1) + c - '0';
    c = getchar();
  }
  x = res * f;
}
const long long N = 200000 + 10;
const int mod = 1e9 + 7;
struct node {
  int nxt[26], fa, len;
  void init() {
    fa = len = 0;
    memset(nxt, 0, sizeof(nxt));
  }
} nod[N];
int last = 1, tot = 1, siz[N];
void add(int w) {
  int p = last;
  int now = last = ++tot;
  nod[tot].init();
  siz[now] = 1;
  nod[now].len = nod[p].len + 1;
  for (; p && !nod[p].nxt[w]; p = nod[p].fa) nod[p].nxt[w] = now;
  if (!p)
    nod[now].fa = 1;
  else {
    int q = nod[p].nxt[w];
    if (nod[q].len == nod[p].len + 1)
      nod[now].fa = q;
    else {
      int nq = ++tot;
      nod[nq] = nod[q];
      nod[nq].len = nod[p].len + 1;
      nod[q].fa = nod[now].fa = nq;
      for (; p && nod[p].nxt[w] == q; p = nod[p].fa) nod[p].nxt[w] = nq;
    }
  }
}
int t;
char s[N];
vector<int> p[N];
void dfs(int u) {
  for (auto to : p[u]) dfs(to), siz[u] += siz[to];
}
int main() {
  read(t);
  while (t--) {
    scanf("%s", s + 1);
    for (int i = 1; i <= tot; i++) siz[i] = 0, p[i].clear();
    tot = last = 1;
    nod[tot].init();
    int n = strlen(s + 1);
    for (int i = 1; i <= n; i++) add(s[i] - 'a');
    for (int i = 2; i <= tot; i++) p[nod[i].fa].push_back(i);
    dfs(1);
    long long ans = 0;
    for (int i = 1; i <= tot; i++) {
      int len = nod[i].len - nod[nod[i].fa].len;
      ans += 1ll * len * siz[i] * siz[i];
    }
    printf("%lld\n", ans);
  }
  return 0;
}