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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return x * f;
}
const int N = 100005;
char s[N];
int t1[N], t2[N], c[N], sa[N], rnk[N], h[N], sk[N], pos[N], n, m = 130;
void getsa() {
  int *x = t1, *y = t2, i, p;
  for (i = 1; i <= m; ++i) c[i] = 0;
  for (i = 1; i <= n; ++i) x[i] = s[i], c[x[i]]++;
  for (i = 1; i <= m; ++i) c[i] += c[i - 1];
  for (i = n; i >= 1; --i) sa[c[x[i]]--] = i;
  for (int k = 1; k <= n; k <<= 1) {
    p = 0;
    for (i = n - k + 1; i <= n; ++i) y[++p] = i;
    for (i = 1; i <= n; ++i)
      if (sa[i] > k) y[++p] = sa[i] - k;
    for (i = 1; i <= m; ++i) c[i] = 0;
    for (i = 1; i <= n; ++i) c[x[y[i]]]++;
    for (i = 1; i <= m; ++i) c[i] += c[i - 1];
    for (i = n; i >= 1; --i) sa[c[x[y[i]]]--] = y[i];
    swap(x, y);
    x[sa[1]] = 1;
    p = 2;
    for (i = 2; i <= n; ++i)
      x[sa[i]] = (y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k])
                     ? p - 1
                     : p++;
    if (p > n) break;
    m = p;
  }
}
void getheight() {
  for (int i = 1; i <= n; ++i) rnk[sa[i]] = i;
  int k = 0;
  h[1] = 0;
  for (int i = 1; i <= n; ++i) {
    if (rnk[i] == 1) continue;
    if (k) k--;
    int j = sa[rnk[i] - 1];
    while (i + k <= n && j + k <= n && s[i + k] == s[j + k]) k++;
    h[rnk[i]] = k;
  }
}
void solve() {
  scanf("%s", s + 1);
  n = strlen(s + 1), m = 130;
  getsa();
  getheight();
  int top = 0, now;
  long long ans = 0, t1, t2;
  for (int i = 2; i <= n + 1; ++i) {
    now = i;
    while (top && h[i] < sk[top]) {
      t1 = i - pos[top] + 1;
      t2 = sk[top] - max(sk[top - 1], h[i]);
      ans += t1 * t1 * t2;
      now = pos[top--];
    }
    while (top && sk[top] == h[i]) now = pos[top--];
    sk[++top] = h[i], pos[top] = now;
  }
  for (int i = 1; i <= n; i++) ans += n - i + 1 - max(h[rnk[i]], h[rnk[i] + 1]);
  cout << ans << "\n";
}
int main() {
  for (int T = read(); T--;) solve();
  return 0;
}