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>
#pragma GCC optimize("O2")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long double eps = 1e-7;
const int inf = 1000000010;
const long long INF = 10000000000000010LL;
const int mod = 1000000007;
const int MAXN = 100010, LOG = 18;
int n, m, k, u, v, x, y, t, a, b;
int Rank[MAXN][LOG], P[MAXN], pw;
int L[MAXN], R[MAXN];
int lcp[MAXN];
string S;
bool cmp(int x, int y) {
  if (Rank[x][pw - 1] != Rank[y][pw - 1])
    return Rank[x][pw - 1] < Rank[y][pw - 1];
  if (max(x, y) + (1 << (pw - 1)) > n) return x > y;
  return Rank[x + (1 << (pw - 1))][pw - 1] < Rank[y + (1 << (pw - 1))][pw - 1];
}
void BuildSuffixArray() {
  for (int i = 1; i <= n; i++) Rank[i][0] = S[i - 1], P[i] = i;
  for (pw = 1; pw < LOG; pw++) {
    sort(P + 1, P + n + 1, cmp);
    Rank[P[1]][pw] = 1;
    for (int i = 2; i <= n; i++)
      Rank[P[i]][pw] = Rank[P[i - 1]][pw] + cmp(P[i - 1], P[i]);
  }
}
int Lcp(int x, int y) {
  int res = 0;
  for (int i = LOG - 1; i >= 0; i--)
    if (Rank[x][i] == Rank[y][i]) {
      x += (1 << i);
      y += (1 << i);
      res |= (1 << i);
    }
  return res;
}
void solve() {
  cin >> S;
  n = S.size();
  BuildSuffixArray();
  for (int i = 1; i < n; i++) lcp[i] = Lcp(P[i], P[i + 1]);
  for (int i = 1; i < n; i++)
    for (L[i] = i - 1; L[i] && lcp[L[i]] >= lcp[i]; L[i] = L[L[i]])
      ;
  for (int i = n - 1; i; i--)
    for (R[i] = i + 1; R[i] < n && lcp[R[i]] > lcp[i]; R[i] = R[R[i]])
      ;
  long long ans = 0;
  for (int i = 1; i < n; i++) ans += 1ll * lcp[i] * (R[i] - i) * (i - L[i]);
  ans = ans * 2 + 1ll * n * (n + 1) / 2;
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T;
  for (cin >> T; T; T--) solve();
  while (clock() < 4975)
    ;
  return 0;
}