//dp_i: i的子树中必须往i祖先上贡献的. (不包含i!!!!!!!!!)
#include<bits/stdc++.h>
using namespace std;
#define IOS {cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);}
#define rep(i,j,k) for(int i=j;i<=k;++i)
#define per(i,j,k) for(int i=j;i>=k;--i)
#define P pair<int,int>
#define ll long long
#define vi vector<int>
const int N=1e6+6;
int fa[N],res[N],dp[N],V,sz[N],ok[N];
vi tu[N];
void dfs(int now){
    int sum=0; sz[now]=1;
    for(int son:tu[now]){
        dfs(son); sum+=dp[son]+1; sz[now]+=sz[son];
    }
    dp[now]=max(0,sum-V);
}
void slv(){
    int n; cin>>n;
    rep(i,1,n){
        tu[i].clear(); res[i]=0;
    }
    rep(i,2,n){
        cin>>fa[i];
        tu[fa[i]].push_back(i);
    }
    int L=1,R=n-1;
    while(L<=R){
        V=L+R>>1; dfs(1);
        if(dp[1])L=V+1; 
        else R=V-1;
    }
    V=L-1; dfs(1); assert(dp[1]);
    V=L;
    ok[1]=(dp[1]==1);
    rep(i,2,n){
        ok[i]=(dp[i]==1 && ok[fa[i]]);
    }
    rep(i,1,n){
        int v=sz[i]-1;
        if(v>V)res[i]=1;
        else if(v<V)res[i]=0;
        else{
            if(ok[i])res[i]=1;
            else res[i]=0;
        }
    }
    rep(i,1,n)cout<<res[i]; cout<<'\n';
}
signed main(){
    IOS
    int T;cin>>T;
    while(T--)slv();
}
/*
1
100000
*/