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 MN = 101234;
int gap, n, rnk[MN], sa[MN], diff[MN], lcp[MN];
string s;
bool Cmp(int a, int b) {
  if (rnk[a] != rnk[b]) return rnk[a] < rnk[b];
  return (a + gap < n && b + gap < n) ? rnk[a + gap] < rnk[b + gap] : a > b;
}
void build() {
  for (int i = 0; i < n; i++) rnk[i] = s[i], sa[i] = i;
  for (gap = 1;; gap *= 2) {
    sort(sa, sa + n, Cmp);
    for (int i = 1; i < n; i++) diff[i] = diff[i - 1] + Cmp(sa[i - 1], sa[i]);
    rnk[sa[0]] = 0;
    for (int i = 1; i < n; i++) rnk[sa[i]] = diff[i];
    if (rnk[sa[n - 1]] == n - 1) return;
  }
}
void LCP() {
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (rnk[i] == n - 1) {
      ans = 0;
      continue;
    }
    int r = rnk[i];
    for (int j = sa[r + 1]; s[i + ans] == s[j + ans];) ans++;
    lcp[r] = ans;
    if (ans) ans--;
  }
}
int main() {
  int q;
  cin >> q;
  while (q--) {
    cin >> s;
    n = s.size();
    s += "$";
    build();
    LCP();
    vector<pair<int, int> > vec;
    long long sum = 0, ans = 2 * n - 1;
    for (int i = 0; i < n - 1; i++) {
      int x = lcp[i], cnt = 1;
      while (vec.size() && vec.back().first > x) {
        sum -= 1ll * vec.back().first * vec.back().second;
        cnt += vec.back().second;
        vec.pop_back();
      }
      vec.push_back({x, cnt});
      sum += 1ll * vec.back().first * vec.back().second;
      ans += 2 * sum + i;
    }
    cout << ans << endl;
  }
}