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 long long B = 27;
const long long mod = 1e9 + 7;
const long long N = 1e5 + 5;
string s;
long long n;
long long ans = 0;
long long sum = 0;
long long hsh[N];
long long power[N];
long long suffix[N];
long long a[N];
void pre() {
  power[0] = 1;
  for (long long i = 1; i < N; i++) {
    power[i] = power[i - 1] * B;
  }
}
void maker() {
  for (long long i = 0; i < n; i++) {
    hsh[i + 1] = hsh[i] * B + (s[i] - 'a' + 1);
  }
}
long long get(long long l, long long r) {
  return hsh[r] - hsh[l] * power[r - l];
}
long long lcp(long long x, long long y) {
  long long l = 0;
  long long r = min(n - x, n - y) + 1;
  while (r - l > 1) {
    long long mid = (l + r) / 2;
    if (get(x, x + mid) == get(y, y + mid)) {
      l = mid;
    } else {
      r = mid;
    }
  }
  return l;
}
bool cmp(long long x, long long y) {
  long long len = lcp(x, y);
  if (len + max(x, y) == n) {
    return x > y;
  }
  return s[len + x] < s[len + y];
}
void solver() {
  cin >> s;
  n = s.size();
  ans = 0;
  sum = 0;
  maker();
  for (long long i = 0; i < n; i++) {
    suffix[i] = i;
  }
  sort(suffix, suffix + n, cmp);
  for (long long i = 0; i < n - 1; i++) {
    a[i] = lcp(suffix[i], suffix[i + 1]);
  }
  vector<long long> stk;
  stk.push_back(-1);
  for (long long i = 0; i < n - 1; i++) {
    while (stk.size() && a[stk.back()] >= a[i]) {
      long long ind = stk.back();
      stk.pop_back();
      sum -= (ind - stk.back()) * a[ind];
    }
    sum += (i - stk.back()) * a[i];
    ans += sum;
    stk.push_back(i);
  }
  cout << 2 * ans + n * (n + 1) / 2 << endl;
}
int main() {
  pre();
  long long t;
  cin >> t;
  for (long long i = 0; i < t; i++) {
    solver();
  }
}
