#include<bits/stdc++.h>
using namespace std;
const int maxn=2e5+10;
typedef long long int ll;
int head[maxn],vis[maxn],fa[maxn];
ll ans[maxn],siz[maxn],w[maxn];
int cnt,n,st,ed,mex;
struct edge{
    int to,next;
}e[maxn*2];
void add(int x,int y){
    e[cnt].to=y;
    e[cnt].next=head[x];
    head[x]=cnt++;
}
void init(int n){
    cnt=0;
    mex=st=ed=1;
    for(int i=0;i<=n;i++){
        ans[i]=vis[i]=siz[i]=0;
        head[i]=-1;
    }
    vis[1]=1;
}
void dfs(int u,int f){
    siz[u]=1;fa[u]=f;
    for(int i=head[u];i!=-1;i=e[i].next){
        int v=e[i].to;
        if(v==f)continue;
        dfs(v,u);
        siz[u]+=siz[v];
    }
}
int main(){
    int t,x,y;
    cin>>t;
    while(t--){
        cin>>n;
        init(n+2);
        for(int i=1;i<n;i++){
            cin>>x>>y;
            x++;y++;
            add(x,y);
            add(y,x);
        }
        dfs(1,0);
        ll res=1;
        ans[1]=n*1ll*(n-1)/2;
        for(int i=head[1];i!=-1;i=e[i].next){
            int v=e[i].to;
            ans[2]+=res*siz[v];res+=siz[v];
        }
        for(int i=2;i<=n;i++){
            if(!vis[i]){
                int now=i;
                while(!vis[now]){
                    if(fa[now]==1 && i==2)siz[1]-=siz[now];
                    vis[now]=1;
                    now=fa[now];
                }
                while(vis[mex])mex++;
                if(st!=now)swap(st,ed);
                if(st!=now)break;
                st=i;
                ans[mex]=siz[st]*siz[ed]; //表示
            }
        }
        ll tmp=0;
        for(int i=n+1;i;i--){
            w[i]=max(0ll,(ans[i]-tmp));
            tmp=max(tmp,ans[i]);
        }
        for(int i=1;i<=n+1;i++)cout<<w[i]<<" ";
        cout<<endl;
    }
    return 0;
}
		  	   	 	 			 			     		
### Prompt

Your task is to create a CPP solution to the following problem:

You are given a tree with n nodes, numerated from 0 to n-1. For each k between 0 and n, inclusive, you have to count the number of unordered pairs (u,v), u ≠ v, such that the MEX of all the node labels in the shortest path from u to v (including end points) is k.

The MEX of a sequence of integers is the smallest non-negative integer that does not belong to the sequence.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases. 

The first line of each test case contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^{5}).

The next n-1 lines of each test case describe the tree that has to be constructed. These lines contain two integers u and v (0 ≤ u,v ≤ n-1) denoting an edge between u and v (u ≠ v).

It is guaranteed that the given edges form a tree.

It is also guaranteed that the sum of n for all test cases does not exceed 2 ⋅ 10^{5}.

Output

For each test case, print n+1 integers: the number of paths in the tree, such that the MEX of all the node labels in that path is k for each k from 0 to n.

Example

Input


2
4
0 1
0 2
2 3
2
1 0


Output


1 2 1 1 1 
0 0 1 

Note

  1. In example case 1, <image>
    * For k = 0, there is 1 path that is from 2 to 3 as MEX([2, 3]) = 0. 
    * For k = 1, there are 2 paths that is from 0 to 2 as MEX([0, 2]) = 1 and 0 to 3 as MEX([0, 2, 3]) = 1. 
    * For k = 2, there is 1 path that is from 0 to 1 as MEX([0, 1]) = 2. 
    * For k = 3, there is 1 path that is from 1 to 2 as MEX([1, 0, 2]) = 3 
    * For k = 4, there is 1 path that is from 1 to 3 as MEX([1, 0, 2, 3]) = 4. 
  2. In example case 2, <image>
    * For k = 0, there are no such paths. 
    * For k = 1, there are no such paths. 
    * For k = 2, there is 1 path that is from 0 to 1 as MEX([0, 1]) = 2. 

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
const int maxn=2e5+10;
typedef long long int ll;
int head[maxn],vis[maxn],fa[maxn];
ll ans[maxn],siz[maxn],w[maxn];
int cnt,n,st,ed,mex;
struct edge{
    int to,next;
}e[maxn*2];
void add(int x,int y){
    e[cnt].to=y;
    e[cnt].next=head[x];
    head[x]=cnt++;
}
void init(int n){
    cnt=0;
    mex=st=ed=1;
    for(int i=0;i<=n;i++){
        ans[i]=vis[i]=siz[i]=0;
        head[i]=-1;
    }
    vis[1]=1;
}
void dfs(int u,int f){
    siz[u]=1;fa[u]=f;
    for(int i=head[u];i!=-1;i=e[i].next){
        int v=e[i].to;
        if(v==f)continue;
        dfs(v,u);
        siz[u]+=siz[v];
    }
}
int main(){
    int t,x,y;
    cin>>t;
    while(t--){
        cin>>n;
        init(n+2);
        for(int i=1;i<n;i++){
            cin>>x>>y;
            x++;y++;
            add(x,y);
            add(y,x);
        }
        dfs(1,0);
        ll res=1;
        ans[1]=n*1ll*(n-1)/2;
        for(int i=head[1];i!=-1;i=e[i].next){
            int v=e[i].to;
            ans[2]+=res*siz[v];res+=siz[v];
        }
        for(int i=2;i<=n;i++){
            if(!vis[i]){
                int now=i;
                while(!vis[now]){
                    if(fa[now]==1 && i==2)siz[1]-=siz[now];
                    vis[now]=1;
                    now=fa[now];
                }
                while(vis[mex])mex++;
                if(st!=now)swap(st,ed);
                if(st!=now)break;
                st=i;
                ans[mex]=siz[st]*siz[ed]; //表示
            }
        }
        ll tmp=0;
        for(int i=n+1;i;i--){
            w[i]=max(0ll,(ans[i]-tmp));
            tmp=max(tmp,ans[i]);
        }
        for(int i=1;i<=n+1;i++)cout<<w[i]<<" ";
        cout<<endl;
    }
    return 0;
}
		  	   	 	 			 			     		
```