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 mod = 1e9 + 7;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const int INF = 0x3f3f3f3f;
const int maxn = 3e5 + 5;
struct NODE {
  int ch[26];
  int len, fa, cnt;
  void clear() {
    memset(ch, 0, sizeof(ch));
    len = 0;
  }
} dian[maxn << 1];
int last, tot;
void init() {
  last = 1;
  tot = 1;
  dian[1].clear();
}
void ins(int c) {
  int p = last, np = last = ++tot;
  dian[np].clear();
  dian[np].cnt = 1;
  dian[np].len = dian[p].len + 1;
  for (; p && !dian[p].ch[c]; p = dian[p].fa) dian[p].ch[c] = np;
  if (!p)
    dian[np].fa = 1;
  else {
    int q = dian[p].ch[c];
    if (dian[q].len == dian[p].len + 1)
      dian[np].fa = q;
    else {
      int nq = ++tot;
      dian[nq].clear();
      dian[nq] = dian[q];
      dian[nq].cnt = 0;
      dian[nq].len = dian[p].len + 1;
      dian[q].fa = dian[np].fa = nq;
      for (; p && dian[p].ch[c] == q; p = dian[p].fa) dian[p].ch[c] = nq;
    }
  }
}
int cnt[maxn << 1], id[maxn << 1];
void bucketsort() {
  memset(cnt, 0, sizeof(cnt));
  for (int i = (1); i < (tot + 1); i++) ++cnt[dian[i].len];
  for (int i = (1); i < (tot + 1); i++) cnt[i] += cnt[i - 1];
  for (int i = (1); i < (tot + 1); i++) id[cnt[dian[i].len]--] = i;
}
void topogao() {
  bucketsort();
  for (int i = (tot + 1) - 1; i >= 1; i--) {
    int u = id[i], fa = dian[u].fa;
    dian[fa].cnt += dian[u].cnt;
  }
}
char s[maxn];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> s;
    int n = strlen(s);
    init();
    for (int i = (0); i < (n); i++) ins(s[i] - 'a');
    topogao();
    long long ans = 0;
    for (int i = (2); i < (tot + 1); i++) {
      int len = dian[i].len - dian[dian[i].fa].len;
      ans += 1ll * len * dian[i].cnt * dian[i].cnt;
    }
    cout << ans << "\n";
  }
  return 0;
}