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 MAXLEN = 100005;
vector<int> graph[2 * MAXLEN];
struct automaton {
  struct state {
    int link;
    int next[27];
    int len;
    long long cnt;
  };
  state st[4 * MAXLEN];
  int sz;
  long long ans = 0;
  void sa_init() {
    ans = 0;
    for (int i = 0; i < sz; i++) {
      memset(st[i].next, 0, sizeof(st[i].next));
      st[i].len = 0;
      st[i].cnt = 0;
      st[i].link = 0;
    }
    sz = 1;
    last = 0;
    st[0].link = -1;
    st[0].len = 0;
    memset(st[0].next, 0, sizeof(st[0].next));
  }
  int last = 0;
  void sa_extend(int c) {
    int cur = sz++;
    st[cur].len = st[last].len + 1;
    st[cur].cnt = 1;
    int p = last;
    while (p != -1 && !st[p].next[c]) {
      st[p].next[c] = cur;
      p = st[p].link;
    }
    if (p == -1) {
      st[cur].link = 0;
    } else {
      int q = st[p].next[c];
      if (st[p].len + 1 == st[q].len) {
        st[cur].link = q;
      } else {
        int clone = sz++;
        for (int i = 0; i < 26; i++) st[clone].next[i] = st[q].next[i];
        st[clone].link = st[q].link;
        st[clone].cnt = 0;
        st[clone].len = st[p].len + 1;
        while (p != -1 && st[p].next[c] == q) {
          st[p].next[c] = clone;
          p = st[p].link;
        }
        st[cur].link = st[q].link = clone;
      }
    }
    last = cur;
  }
  void pre() {
    for (int i = 0; i < sz; i++) {
      graph[i].clear();
    }
    for (int i = 1; i < sz; i++) {
      graph[st[i].link].push_back(i);
    }
  }
  void dfs(int u) {
    for (auto &x : graph[u]) {
      dfs(x);
      st[u].cnt += st[x].cnt;
    }
    ans += (1LL * st[u].cnt * st[u].cnt) * (st[u].len - st[st[u].link].len);
  }
} a;
char s[MAXLEN];
int main() {
  int test;
  scanf("%d", &test);
  while (test--) {
    scanf("%s", s);
    int n = strlen(s);
    a.sa_init();
    for (int i = 0; i < n; i++) {
      a.sa_extend(s[i] - 'a');
    }
    a.pre();
    a.dfs(0);
    printf("%lld\n", a.ans);
  }
  return 0;
}