#include<bits/stdc++.h>
#define ll long long

using namespace std;
const int N = 1e5 + 10;
struct Magic {
    vector<int>neg, pos;
    int z, sum;
    Magic() : z(0), sum(0) {};
    Magic(vector<int>& v) : z(0), sum(0) {
        int i = 0, j = (int)v.size() - 1;
        while (i < v.size() && v[i] < 0)neg.push_back(v[i]), sum += v[i], i++;
        while (j >= 0 && v[j] > 0)pos.push_back(v[j]), j--;
        z = j - i + 1;
    };

    int size()
    {
        return neg.size() + pos.size() + z;
    }

    int data(int i)
    {
        if (i < neg.size())return neg[i];
        else if (i < neg.size() + z)return 0;
        else return pos[pos.size() - (i - z - neg.size()) - 1];
    }

    void insert(int x)
    {
        if (x < 0)neg.push_back(x), sum += x;
        else pos.push_back(x);
    }
};
int n, ans[N];
vector<int>e[N];
string s;

/*
总之对于突序列的维护就是维护首项和他们的差分数组
闵可夫斯基告诉我们差分数组直接归并为有序数组即可
那么首项呢
*/

//超时大概就是我们合并写错了

pair<int, Magic> dfs(int u)
{
    Magic magic;
    int val = 0;//这就是首项
    for (auto v : e[u])
    {
        auto res = dfs(v);
        if (magic.size() == 0)magic = move(res.second), val += res.first;
        else {
            int len = min(magic.size(), res.second.size());
            //暴力合并
            val += res.first;
            vector<int>v(len);
            for (int i = 0; i < len; i++)v[i] = magic.data(i) + res.second.data(i);
            magic = Magic(v);
        }
    }
    //闵可夫斯基和
    if (s[u - 1] == '0')// 1
    {
        magic.insert(1);
    }
    else {//-1
        val++, magic.insert(-1);
    }
    ans[u] = val + magic.sum;

    return pair<int, Magic>(val, magic);
}
void solve()
{
    cin >> n >> s;
    for (int i = 1; i <= n; i++)e[i].clear();
    for (int i = 2; i <= n; i++) {
        int p; cin >> p;
        e[p].push_back(i);
    }
    dfs(1);
    for (int i = 1; i <= n; i++)cout << ans[i] << ' ';
    cout << '\n';
}

int main()
{
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int t; cin >> t;
    while (t--)solve();
    return 0;
}