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 maxn = 100010;
const int maxsiz = 130;
long long T, n, m, sa[maxn], x[maxn], y[maxn], c[maxn], rnk[maxn], h[maxn], ans,
    tot, cnt;
char s[maxn];
struct node {
  long long v, cnt;
};
stack<node> st;
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%s", s + 1);
    n = strlen(s + 1);
    m = maxsiz - 1;
    for (int i = 1; i <= n; i++) x[i] = s[i];
    for (int i = 0; i <= m; i++) c[i] = 0;
    for (int i = 1; i <= n; i++) c[x[i]]++;
    for (int i = 1; i <= m; i++) c[i] += c[i - 1];
    for (int i = n; i >= 1; i--) sa[c[x[i]]--] = i;
    for (int k = 1, cur; k <= n; k <<= 1) {
      cur = 0;
      for (int i = n; i > n - k; i--) y[++cur] = i;
      for (int i = 1; i <= n; i++)
        if (sa[i] > k) y[++cur] = sa[i] - k;
      for (int i = 0; i <= m; i++) c[i] = 0;
      for (int i = 1; i <= n; i++) c[x[i]]++;
      for (int i = 1; i <= m; i++) c[i] += c[i - 1];
      for (int i = n; i >= 1; i--) sa[c[x[y[i]]]--] = y[i];
      cur = y[sa[1]] = 1;
      for (int i = 2; i <= n; i++) {
        int a, b;
        a = sa[i] + k > n ? -1 : x[sa[i] + k];
        b = sa[i - 1] + k > n ? -1 : x[sa[i - 1] + k];
        y[sa[i]] = (x[sa[i]] == x[sa[i - 1]]) && (a == b) ? cur : (++cur);
      }
      swap(x, y);
      if (cur >= n) break;
      m = cur;
    }
    for (int i = 1; i <= n; i++) rnk[sa[i]] = i;
    int nww = 0;
    for (int i = 1; i <= n; i++) {
      if (rnk[i] == 1) continue;
      if (nww) nww--;
      while (i + nww <= n && sa[rnk[i] - 1] + nww <= n &&
             s[i + nww] == s[sa[rnk[i] - 1] + nww])
        nww++;
      h[rnk[i]] = nww;
    }
    ans = n * (n + 1) / 2;
    for (int i = 1; i <= n; i++) {
      cnt = 0;
      while (!st.empty() && st.top().v >= h[i])
        cnt += st.top().cnt, tot -= st.top().v * st.top().cnt, st.pop();
      st.push({h[i], cnt + 1});
      tot += h[i] * (cnt + 1);
      ans += 2 * tot;
    }
    printf("%lld\n", ans);
  }
  return 0;
}