题意：https://www.luogu.com.cn/problem/CF1678B2

思路：对于这种每段都是偶数的题来说，我们直接划分成最小的偶数段也就是2个一段，然后出现不一样我们就需要操作一次，对于最小段数来说我们直接看有多少已经确定的段数就行。

/*keep on going and never give up*/
#include<cstdio>
#include<iostream>
#include<queue>
#include<algorithm>
using namespace std;
#define int long long
typedef pair<int, int> pii;
#define lowbit(x) x&(-x)
#define endl '\n'
#define wk is zqx ta die
signed main() {
	std::ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int t;
	cin >> t;
	while (t--) {
		int n;
		cin >> n;
		string s;
		cin >> s;
		int ans = 0;
		int ans1 = 0;
		int now = -1;
		for (int i = 0; i < s.size(); i += 2) {
			if (s[i] == s[i + 1] && (s[i] - '0') != now) {
				now = s[i] - '0';
				ans++;
			}
			if (s[i] != s[i + 1]) {
				ans1++;
			}
		}
		if (ans == 0) {
			ans++;
		}
		cout << ans1 << " " << ans << endl;
	}
	return 0;
}