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 MAXP = 17;
char S[100005];
int N;
pair<pair<int, int>, int> T[100005];
int sufrank[MAXP + 1][100005];
void getSuffixArray() {
  for (int i = 0; i < N; ++i) sufrank[0][i] = S[i];
  for (int l = 1; l <= MAXP; ++l) {
    int len = 1 << (l - 1);
    for (int i = 0; i < N; ++i) {
      int nex = i + len < N ? sufrank[l - 1][i + len] : -1;
      T[i] = {{sufrank[l - 1][i], nex}, i};
    }
    sort(T, T + N);
    for (int i = 0; i < N; ++i) {
      if (i && T[i].first == T[i - 1].first)
        sufrank[l][T[i].second] = sufrank[l][T[i - 1].second];
      else
        sufrank[l][T[i].second] = i;
    }
  }
}
int getLcp(int a, int b) {
  int len = 0;
  for (int l = MAXP; l >= 0; --l) {
    if (a + len >= N || b + len >= N) break;
    if (sufrank[l][a + len] == sufrank[l][b + len]) len += 1 << l;
  }
  return len;
}
int P[100005];
void solve() {
  scanf("%s", S);
  N = strlen(S);
  getSuffixArray();
  long long ans = 0;
  for (int i = 0; i < N; ++i) {
    ans += N - i;
    P[sufrank[MAXP][i]] = i;
  }
  long long add = 0;
  stack<pair<int, int>> st;
  for (int i = 1; i < N; ++i) {
    int gap = getLcp(P[i], P[i - 1]);
    int cnt = 1;
    while (!st.empty() && st.top().first >= gap) {
      cnt += st.top().second;
      add -= (long long)st.top().second * st.top().first;
      st.pop();
    }
    st.push({gap, cnt});
    add += (long long)gap * cnt;
    ans += add * 2;
  }
  printf("%lld\n", ans);
}
int main() {
  int TC;
  scanf("%d", &TC);
  while (TC--) {
    solve();
  }
}