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;
inline int read() {
  int ans = 0;
  char ch = getchar();
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) ans = ((ans << 2) + ans << 1) + (ch ^ 48), ch = getchar();
  return ans;
}
const int mod = 1e9 + 7;
inline int add(int a, int b) { return (a += b) < mod ? a : a - mod; }
inline int dec(int a, int b) { return (a -= b) < 0 ? a + mod : a; }
inline int mul(int a, int b) { return (long long)a * b % mod; }
inline void Add(int& a, int b) { (a += b) < mod ? a : (a -= mod); }
inline void Dec(int& a, int b) { (a -= b) < 0 ? (a += mod) : a; }
inline void Mul(int& a, int b) { a = (long long)a * b % mod; }
inline int ksm(int a, int p) {
  int ret = 1;
  for (; p; p >>= 1, Mul(a, a))
    if (p & 1) Mul(ret, a);
  return ret;
}
const int N = 10005;
char s[N];
int n;
int main() {
  for (register int tt = read(); tt; --tt) {
    n = read(), scanf("%s", s + 1);
    int pre = 0, suf = 0;
    for (register int i = 1; i <= n; ++i) {
      if (s[i] == '<')
        ++pre;
      else
        break;
    }
    for (register int i = n; i; --i) {
      if (s[i] == '>')
        ++suf;
      else
        break;
    }
    cout << min(pre, suf) << '\n';
  }
  return 0;
}
