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.
#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 5;
char s[N];
int n, m, tax[N], rnk[N], sa[N], h[N], tp[N];
void Sort() {
  for (int i = 0; i <= m; ++i) tax[i] = 0;
  for (int i = 1; i <= n; ++i) tax[rnk[i]]++;
  for (int i = 1; i <= m; ++i) tax[i] += tax[i - 1];
  for (int i = n; i >= 1; --i) sa[tax[rnk[tp[i]]]--] = tp[i];
}
void SuffixSort() {
  m = 128;
  for (int i = 1; i <= n; ++i) rnk[i] = s[i] - '0' + 1, tp[i] = i;
  Sort();
  for (int w = 1, p = 0; p < n; m = p, w <<= 1) {
    p = 0;
    for (int i = 1; i <= w; ++i) tp[++p] = n - w + i;
    for (int i = 1; i <= n; ++i)
      if (sa[i] > w) tp[++p] = sa[i] - w;
    Sort();
    memcpy(tp, rnk, sizeof(rnk));
    rnk[sa[1]] = p = 1;
    for (int i = 2; i <= n; ++i)
      rnk[sa[i]] =
          (tp[sa[i]] == tp[sa[i - 1]] && tp[sa[i] + w] == tp[sa[i - 1] + w])
              ? p
              : ++p;
  }
  int k = 0, j;
  for (int i = 1; i <= n; ++i) {
    if (k) --k;
    j = sa[rnk[i] - 1];
    while (s[i + k] == s[j + k]) ++k;
    h[rnk[i]] = k;
  }
}
int st[N], L[N], R[N], c[N];
int top = 0;
signed main() {
  int T = 1;
  cin >> T;
  while (T-- > 0) {
    scanf("%s", s + 1);
    n = strlen(s + 1);
    SuffixSort();
    long long ans = 1ll * n * (n + 1) / 2, cnt, tot = 0;
    int j;
    top = 0;
    for (int i = 1; i <= n; ++i) {
      cnt = 1;
      while (top && h[st[top]] >= h[i]) {
        j = st[top];
        tot -= 1ll * c[j] * h[j];
        cnt += c[j];
        top--;
      }
      ++top;
      st[top] = i;
      c[i] = cnt;
      tot += 1ll * h[i] * cnt;
      ans += 2ll * tot;
    }
    cout << ans << "\n";
  }
  return 0;
}
