You have a string s of length n consisting of only characters > and <. You may do some operations with this string, for each operation you have to choose some character that still remains in the string. If you choose a character >, the character that comes right after it is deleted (if the character you chose was the last one, nothing happens). If you choose a character <, the character that comes right before it is deleted (if the character you chose was the first one, nothing happens).

For example, if we choose character > in string > > < >, the string will become to > > >. And if we choose character < in string > <, the string will become to <.

The string is good if there is a sequence of operations such that after performing it only one character will remain in the string. For example, the strings >, > > are good. 

Before applying the operations, you may remove any number of characters from the given string (possibly none, possibly up to n - 1, but not the whole string). You need to calculate the minimum number of characters to be deleted from string s so that it becomes good.

Input

The first line contains one integer t (1 ≤ t ≤ 100) – the number of test cases. Each test case is represented by two lines.

The first line of i-th test case contains one integer n (1 ≤ n ≤ 100) – the length of string s.

The second line of i-th test case contains string s, consisting of only characters > and <.

Output

For each test case print one line.

For i-th test case print the minimum number of characters to be deleted from string s so that it becomes good.

Example

Input


3
2
&lt;&gt;
3
&gt;&lt;&lt;
1
&gt;


Output


1
0
0

Note

In the first test case we can delete any character in string <>.

In the second test case we don't need to delete any characters. The string > < < is good, because we can perform the following sequence of operations: > < < → < < → <.
#include <bits/stdc++.h>
using namespace std;
void swap(int *a, int *b) {
  int temp = *a;
  *a = *b;
  *b = temp;
}
int isprime(int n) {
  if (n < 2) return 0;
  if (n < 4) return 1;
  if (n % 2 == 0 or n % 3 == 0) return 0;
  for (int i = 5; i * i <= n; i += 6)
    if (n % i == 0 or n % (i + 2) == 0) return 0;
  return 1;
}
void pairsort(int a[], int b[], int n) {
  pair<int, int> pairt[n];
  for (int i = 0; i < n; i++) {
    pairt[i].first = a[i];
    pairt[i].second = b[i];
  }
  sort(pairt, pairt + n);
  for (int i = 0; i < n; i++) {
    a[i] = pairt[i].first;
    b[i] = pairt[i].second;
  }
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long fact(long long a) {
  long long p = 1;
  for (int i = 1; i <= a; i++) p *= i;
  return p;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int q;
  cin >> q;
  while (q--) {
    int n;
    cin >> n;
    string s;
    cin >> s;
    int x = 0, y = 0, flag = 0;
    int i = 0;
    while (s[i] == '<' && i < n) {
      x++;
      i++;
    }
    i = n - 1;
    while (s[i] == '>' && i >= 0) {
      y++;
      i--;
    }
    cout << min(x, y) << "\n";
  }
  return 0;
}
