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;
template <typename T, typename U>
inline void chkmax(T& x, U y) {
  if (x < y) x = y;
}
template <typename T, typename U>
inline void chkmin(T& x, U y) {
  if (y < x) x = y;
}
const int MAXN = 222222;
char s[MAXN];
int L[MAXN], C[MAXN], E[MAXN][26], P[MAXN], ID[MAXN], cnt[MAXN];
int n, m, cur;
void extend(int c) {
  int x = cur, y = cur = ++m;
  L[cur] = L[x] + 1, C[cur] = 1;
  for (; x && !E[x][c]; x = P[x]) E[x][c] = y;
  if (!x)
    P[y] = 1;
  else {
    int z = E[x][c];
    if (L[z] == L[x] + 1)
      P[y] = z;
    else {
      int u = ++m;
      L[u] = L[x] + 1, memcpy(E[u], E[z], sizeof(E[u])), P[u] = P[z],
      P[z] = P[y] = u;
      for (; x && E[x][c] == z; x = P[x]) E[x][c] = u;
    }
  }
}
int main() {
  int ncase;
  for (scanf("%d", &ncase); ncase--;) {
    memset(E, 0, sizeof E);
    memset(L, 0, sizeof L);
    memset(P, 0, sizeof P);
    memset(C, 0, sizeof C);
    m = cur = 1;
    scanf("%s", s);
    n = strlen(s);
    for (int i = 0; s[i]; i++) extend(s[i] - 'a');
    memset(cnt, 0, sizeof cnt);
    for (int i = 1; i <= m; i++) cnt[L[i]]++;
    for (int i = 1; i <= n; i++) cnt[i] += cnt[i - 1];
    for (int i = m; i > 0; i--) ID[cnt[L[i]]--] = i;
    for (int i = m, u; i > 0; i--) u = ID[i], C[P[u]] += C[u];
    unsigned long long ans = 0;
    for (int i = 1; i <= m; i++) ans += 1ULL * C[i] * C[i] * (L[i] - L[P[i]]);
    cout << ans << endl;
  }
  return 0;
}