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;
long long powermodm(long long x, long long n, long long M) {
  long long result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % M;
    x = (x * x) % M;
    n = n / 2;
  }
  return result;
}
long long power(long long _a, long long _b) {
  long long _r = 1;
  while (_b) {
    if (_b % 2 == 1) _r = (_r * _a);
    _b /= 2;
    _a = (_a * _a);
  }
  return _r;
}
long long isPrime[100001];
void sieve();
vector<long long> Divisors(long long n);
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    string a;
    cin >> a;
    long long cnt1 = 0, cnt2 = 0;
    for (long long i = 0; i < n; i++) {
      if (a[i] == '>')
        cnt1++;
      else
        cnt2++;
    }
    long long ans, cn1 = 0, cn2 = 0;
    string b;
    b = a;
    long long hi = n;
    for (long long i = 0; i < n; i++) {
      if (a[i] == '>' && i != (n - 1)) {
        if (a[i + 1] == '<') {
          cn1++;
          a.erase(a.begin() + i + 1);
          n--;
          i = i - 2;
        }
      }
    }
    ans = cnt2 - cn1;
    a = b;
    n = hi;
    for (long long i = 0; i < n; i++) {
      if (a[i] == '<' && i != 0) {
        if (a[i - 1] == '>') {
          cn2++;
          a.erase(a.begin() + i - 1);
          n--;
          i = i - 2;
        }
      }
    }
    long long hey = cnt1 - cn2;
    ans = min(ans, hey);
    cout << ans << endl;
  }
}
