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;
struct state {
  int len, link;
  map<char, int> next;
};
const int MAXN = 1 << 21;
state st[MAXN];
int sz, last;
int terminal[MAXN];
int occ[MAXN];
void sa_init() {
  st[0].len = 0;
  st[0].link = -1;
  ++sz;
  last = 0;
}
void sa_extend(char c) {
  int cur = sz++;
  st[cur].len = st[last].len + 1;
  int p = last;
  while (p != -1 && !st[p].next.count(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++;
      st[clone].len = st[p].len + 1;
      st[clone].next = st[q].next;
      st[clone].link = st[q].link;
      while (p != -1 && st[p].next[c] == q) {
        st[p].next[c] = clone;
        p = st[p].link;
      }
      st[q].link = st[cur].link = clone;
    }
  }
  last = cur;
}
void dfs(int v) {
  if (occ[v] != 0) return;
  occ[v] = terminal[v];
  for (auto e : st[v].next) {
    dfs(e.second);
    occ[v] += occ[e.second];
  }
}
vector<int> g[MAXN];
void calc(int v) {
  if (terminal[v] != 0) return;
  for (auto x : g[v]) {
    if (x == 0) {
      terminal[v] += 1;
      continue;
    }
    if (terminal[x] == 0) {
      calc(x);
    }
    terminal[v] += terminal[x];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    for (int i = 0; i < (int)sz; ++i) {
      st[i] = state();
      occ[i] = terminal[i] = 0;
      g[i].clear();
    }
    sz = 0;
    string s;
    cin >> s;
    sa_init();
    int n = s.size();
    for (int i = 0; i < (int)n; ++i) {
      sa_extend(s[i]);
    }
    int v = last;
    while (v > 0) {
      terminal[v] = 1;
      v = st[v].link;
    }
    dfs(0);
    v = -1;
    for (int i = 0; i < (int)sz; ++i) {
      if (st[i].next.empty()) {
        v = i;
      }
      for (auto x : st[i].next) {
        g[x.second].push_back(i);
      }
      terminal[i] = 0;
    }
    calc(v);
    long long ans = 0;
    for (int i = 0; i < (int)sz; ++i) {
      ans += (1LL * terminal[i] * occ[i] * occ[i]);
    }
    cout << ans << '\n';
  }
}