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;
const int mod = 1e9;
long long xor1_n(long long n) {
  switch (n % 4) {
    case 0:
      return n;
    case 1:
      return 1;
    case 2:
      return n + 1;
  }
  return 0;
}
int ncr(int n, int r) {
  int res = 1;
  if (r > n - r) r = n - r;
  for (int i = 0; i < r; i++) {
    res *= n - i;
    res /= i + 1;
  }
  return res;
}
clock_t time_p = clock();
void rtime() {
  time_p = clock() - time_p;
  cerr << "Time Taken : " << (float)(time_p) / CLOCKS_PER_SEC << "\n";
}
long long power(long long a, long long b) {
  long long ans = 1;
  a = a % mod;
  while (b > 0) {
    if (b & 1) ans = (1ll * a * ans) % mod;
    b >>= 1;
    a = (1ll * a * a) % mod;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  cin >> t;
  while (t--) {
    int n;
    string s;
    cin >> n >> s;
    int pos;
    for (int i = 0; i < n; i++) {
      if (s[i] == '>') {
        pos = i;
        break;
      }
    }
    int x;
    for (int i = n - 1; i >= 0; i--) {
      if (s[i] == '<') {
        x = n - i - 1;
        break;
      }
    }
    int ans = min(pos, x);
    cout << ans << "\n";
  }
  rtime();
  return 0;
}
