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;
void file(string s) {
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
template <typename Tp>
void read(Tp &x) {
  int fh = 1;
  char c = getchar();
  x = 0;
  while (c > '9' || c < '0') {
    if (c == '-') {
      fh = -1;
    }
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + (c & 15);
    c = getchar();
  }
  x *= fh;
}
int n, m;
long long ans;
namespace SAM {
int ch[2000005][26], mxl[2000005], fail[2000005], tot, lst;
int edp[2000005];
vector<int> ft[2000005];
void clear() {
  for (int i = 1; i <= tot; ++i)
    memset(ch[i], 0, sizeof(ch[i])), mxl[i] = fail[i] = edp[i] = 0,
                                     ft[i].clear();
  tot = lst = 1;
  ans = 0;
}
void reset() { lst = 1; }
void add_char(int c) {
  int p = lst, np = ++tot;
  lst = tot;
  edp[tot] = 1;
  mxl[np] = mxl[p] + 1;
  for (; p && !ch[p][c]; p = fail[p]) ch[p][c] = np;
  if (!p)
    fail[np] = 1;
  else {
    int x = ch[p][c];
    if (mxl[x] == mxl[p] + 1)
      fail[np] = x;
    else {
      int nx = ++tot;
      memcpy(ch[nx], ch[x], sizeof(ch[x]));
      fail[nx] = fail[x];
      mxl[nx] = mxl[p] + 1;
      for (; p && ch[p][c] == x; p = fail[p]) ch[p][c] = nx;
      fail[x] = fail[np] = nx;
    }
  }
}
void ins(const char *str) {
  reset();
  int nn = strlen(str);
  for (int i = 0; i < nn; ++i) add_char(str[i] - 'a');
}
void dfs(int x) {
  for (unsigned i = 0; i < ft[x].size(); ++i) {
    int y = ft[x][i];
    dfs(y);
    edp[x] += edp[y];
  }
}
void solve() {
  for (int i = 2; i <= tot; ++i) ft[fail[i]].push_back(i);
  dfs(1);
  for (int i = 1; i <= tot; ++i) {
    ans += 1ll * edp[i] * edp[i] * (mxl[i] - mxl[fail[i]]);
  }
}
}  // namespace SAM
char sstr[2000005];
signed main() {
  int CasT;
  read(CasT);
  while (CasT--) {
    SAM::clear();
    scanf("%s", sstr);
    SAM::ins(sstr);
    SAM::solve();
    printf("%lld\n", ans);
  }
  return 0;
}