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 N = 2e5 + 77;
int a[N], n, k;
char st[N];
struct SAM {
  int las = 1, tot = 1, ch[N][30], fail[N], ml[N], p[N], c[N], siz[N];
  long long ans;
  void clear() {
    las = tot = 1;
    memset(ch, 0, sizeof(ch)), memset(fail, 0, sizeof(fail)),
        memset(ml, 0, sizeof(ml));
    memset(a, 0, sizeof(a)), memset(siz, 0, sizeof(siz));
    memset(c, 0, sizeof(c));
    memset(p, 0, sizeof(p));
  }
  void ins(int c) {
    int p = las, np = las = ++tot;
    ml[np] = ml[p] + 1;
    siz[np] = 1;
    while (p && !ch[p][c]) ch[p][c] = np, p = fail[p];
    if (!p)
      fail[np] = 1;
    else {
      int q = ch[p][c];
      if (ml[q] == ml[p] + 1)
        fail[np] = q;
      else {
        int nq = ++tot;
        ml[nq] = ml[p] + 1;
        for (int i = 0; i < 26; i++) ch[nq][i] = ch[q][i];
        fail[nq] = fail[q];
        fail[q] = fail[np] = nq;
        while (p && ch[p][c] == q) ch[p][c] = nq, p = fail[p];
      }
    }
  }
  void work() {
    for (int i = 1; i <= tot; i++) c[i] = p[i] = 0;
    for (int i = 1; i <= tot; i++) c[ml[i]]++;
    for (int i = 1; i <= tot; i++) c[i] += c[i - 1];
    for (int i = 1; i <= tot; i++) p[c[ml[i]]--] = i;
    ans = 0;
    for (int i = tot; i >= 1; i--) {
      int u = p[i];
      siz[fail[u]] += siz[u];
      ans += 1ll * siz[u] * siz[u] * (ml[u] - ml[fail[u]]);
    }
    printf("%lld\n", ans);
  }
} sam;
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%s", st + 1);
    n = strlen(st + 1);
    sam.clear();
    for (int i = 1; i <= n; i++) sam.ins(st[i] - 'a');
    sam.work();
  }
}