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// STADARD HEADERS
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>

using namespace std;

typedef vector<int> VI;

#define FOR(x, b, e) for (int x = b; x <= (e); ++x)
#define FORD(x, b, e) for (int x = b; x >= (e); --x)
#define REP(x, n) for (int x = 0; x < (n); ++x)
#define VAR(v, n) __typeof(n) v = (n)
#define ALL(c) (c).begin(), (c).end()
#define SIZE(x) ((int)(x).size())
#define FOREACH(i, c) for (VAR(i, (c).begin()); i != c.end(); ++i)
#define PB push_back
#define ST first
#define ND second
#define MP(a, b) make_pair(a, b)
#define INF 2147483000
// STADARD HEADERS

priority_queue<int> pque;

void Solver()
{
    int n;
    int t = 0;
    int St = -1, Fn = -1;
    int ans = 0;
    string s;

    cin >> n;
    cin >> s;

    for(int i = 0; i < n; i++) // od poczatku
    {
        if(s[i] == '1')
        {
            St = i;
            break;
        }
    }

    for(int i = n-1; i >= St-1; i--) // od konca
    {
        if(s[i] == '1')
        {
            Fn = n-1 - i;
            break;
        }
    }

    if(St == Fn && St == -1)
    {
        cout << 0 << "\n";
        return;
    }

    // Find potetential breaks
    int counter = 0;
    for(int i = max(St+1, 0); i <= min(n-Fn, n-1); i++)
    {
        if(s[i] == '0')
        {
            counter++;
            continue;
        }
        else if(s[i] == '1')
        {
            pque.push(counter);
            counter = 0;
        }
    }

    while(pque.top() - t*2 > 0 || St - t > 0 || Fn - t > 0)
    {
        if(pque.top() - t*2 < St-t)
        {
            ans += St-t;
            St = -1;
            t += 1;
            continue;
        }
        if(pque.top() - t*2 < Fn-t)
        {
            ans += Fn-t;
            Fn = -1;
            t += 1;
            continue;
        }

        ans += pque.top() - t*2 - 1;
        pque.pop();
        t += 2;
    }

    cout << n-ans << "\n";
}

int main()
{
    // FAST IO
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int t;
    cin >> t;

    REP(i, t)
        Solver();
}

/*
3
8
00110100
10
1001000010
4
0000
*/