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 inf = 1e9 + 7;
string to_string(string s) { return '"' + s + '"'; }
string to_string(char s) { return string(1, s); }
string to_string(const char *s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A>
string to_string(A);
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool f = 1;
  string r = "{";
  for (const auto &x : v) {
    if (!f) r += ", ";
    f = 0;
    r += to_string(x);
  }
  return r + "}";
}
void debug_out() { cout << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cout << " " << to_string(H);
  debug_out(T...);
}
struct SuffixArray {
  const int L;
  string s;
  vector<vector<int> > P;
  vector<pair<pair<int, int>, int> > M;
  SuffixArray(const string &s)
      : L(s.length()), s(s), P(1, vector<int>(L, 0)), M(L) {
    for (int i = 0; i < L; i++) P[0][i] = int(s[i]);
    for (int skip = 1, level = 1; skip < L; skip *= 2, level++) {
      P.push_back(vector<int>(L, 0));
      for (int i = 0; i < L; i++)
        M[i] =
            make_pair(make_pair(P[level - 1][i],
                                i + skip < L ? P[level - 1][i + skip] : -1000),
                      i);
      sort(M.begin(), M.end());
      for (int i = 0; i < L; i++)
        P[level][M[i].second] = (i > 0 && M[i].first == M[i - 1].first)
                                    ? P[level][M[i - 1].second]
                                    : i;
    }
  }
  vector<int> GetSuffixArray() { return P.back(); }
  vector<int> GetInvSuffixArray() {
    vector<int> ans(int(P.back().size()));
    for (int i = 0; i < int(P.back().size()); ++i) ans[P.back()[i]] = i;
    return ans;
  }
  int LongestCommonPrefix(int i, int j) {
    int len = 0;
    if (i == j) return L - i;
    for (int k = P.size() - 1; k >= 0 && i < L && j < L; k--) {
      if (P[k][i] == P[k][j]) {
        i += 1 << k;
        j += 1 << k;
        len += 1 << k;
      }
    }
    return len;
  }
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    SuffixArray SA(s);
    int n = int(s.size());
    long long ans = n * (n + 1LL) / 2;
    vector<int> P(n - 1);
    auto bc = SA.GetInvSuffixArray();
    for (auto i = (1); i <= (int(bc.size()) - 1); ++i) {
      P[i - 1] = SA.LongestCommonPrefix(bc[i - 1], bc[i]);
    }
    vector<int> L(n - 1, -1), R(n - 1, n - 1);
    stack<int> st;
    for (auto i = (0); i <= (int(P.size()) - 1); ++i) {
      while (!st.empty() and P[st.top()] > P[i]) st.pop();
      if (!st.empty()) L[i] = st.top();
      st.push(i);
    }
    while (!st.empty()) st.pop();
    for (auto i = (int(P.size()) - 1); i >= (0); --i) {
      while (!st.empty() and P[st.top()] >= P[i]) st.pop();
      if (!st.empty()) R[i] = st.top();
      st.push(i);
    }
    for (auto i = (0); i <= (int(P.size()) - 1); ++i) {
      ans += 2 * P[i] * (R[i] - i) * 1LL * (i - L[i]);
    }
    cout << ans << "\n";
  }
  return 0;
}