#include <bits/stdc++.h>
#define endl '\n'
#define inf 100000000000
#define ll long long
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define loop(i,a,b) for(ll i=a;i<b;i++)
#define rloop(i,a,b) for(ll i=a;i>b;i--)


using namespace std;
ll n;
vector<vector<pair<ll,ll>>> tree(200005);
ll ans[200005],parent[200005],parent1[200005];
ll dfs(ll root,ll par)
{
    parent[root]=par;
    ll sum=0;
    for(auto &i:tree[root])
    {
        if(i.fi==par) continue;
        i.se=dfs(i.fi,root);
        sum+=i.se;
    }
    return sum+1;
}
ll find(ll a)
{
    if(parent1[a]==0) return a;
    ll x=find(parent[a]);
    parent1[a]=0;
    return x;
}
void solve()
{
    ll n;
    cin>>n;
    for(ll i=0;i<n;i++) 
    {
        tree[i].clear(); 
        parent1[i]=-1;
    }
    for(ll i=0;i<n-1;i++)
    {
        ll u,v;
        cin>>u>>v;
        tree[u].pb({v,-1});
        tree[v].pb({u,-1});
    }
    dfs(0,-1);
    ans[0]=(n*(n-1))/2;
    ll sum=1,last=0,f=1,head=0,tail=0,x1,curr=1;
    parent1[0]=0;
    while(curr)
    {
        x1=curr;
        curr=parent[curr];
    }
    curr=1;
    for(auto i:tree[0])
    {
        last+=i.se*sum;
        sum+=i.se;
        if(i.fi!=x1) curr+=i.se;
    }
    ans[0]-=last;
    for(ll i=1;i<n;i++)
    {
        ll x=find(i);
        if(x==i)
        {
            ans[i]=0;
            continue;
        }
        if(x==head)
        {
            ll sum1=1,sum2=1;
            for(auto j:tree[i])
            {
                if(j.fi!=parent[i]) sum1+=j.se;
            }
            if(tail){
            for(auto j:tree[tail])
            {
                if(j.fi!=parent[tail]) sum2+=j.se;
            }}
            else sum2=curr;
            ans[i]=last-(sum1*sum2);
            last=sum1*sum2;
            head=i;
        }
        else if(x==tail)
        {
            ll sum1=1,sum2=1;
            for(auto j:tree[i])
            {
                if(j.fi!=parent[i]) sum1+=j.se;
            }
            if(head){
            for(auto j:tree[head])
            {
                if(j.fi!=parent[head]) sum2+=j.se;
            }}
            else sum2=curr;
            ans[i]=last-(sum1*sum2);
            last=sum1*sum2;
            tail=i;
        }
        else
        {
           f=0;
           ans[i]=last;
           for(ll j=i+1;j<=n;j++) ans[j]=0;
           break;
        }
    }
    if(f) ans[n]=1;
    for(ll i=0;i<=n;i++) cout<<ans[i]<<" ";
    cout<<endl;
}

int main() {
    ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	ll t=1;
	cin>>t;
	while(t--)
	{
	   solve();
	}
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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>
#define endl '\n'
#define inf 100000000000
#define ll long long
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define loop(i,a,b) for(ll i=a;i<b;i++)
#define rloop(i,a,b) for(ll i=a;i>b;i--)


using namespace std;
ll n;
vector<vector<pair<ll,ll>>> tree(200005);
ll ans[200005],parent[200005],parent1[200005];
ll dfs(ll root,ll par)
{
    parent[root]=par;
    ll sum=0;
    for(auto &i:tree[root])
    {
        if(i.fi==par) continue;
        i.se=dfs(i.fi,root);
        sum+=i.se;
    }
    return sum+1;
}
ll find(ll a)
{
    if(parent1[a]==0) return a;
    ll x=find(parent[a]);
    parent1[a]=0;
    return x;
}
void solve()
{
    ll n;
    cin>>n;
    for(ll i=0;i<n;i++) 
    {
        tree[i].clear(); 
        parent1[i]=-1;
    }
    for(ll i=0;i<n-1;i++)
    {
        ll u,v;
        cin>>u>>v;
        tree[u].pb({v,-1});
        tree[v].pb({u,-1});
    }
    dfs(0,-1);
    ans[0]=(n*(n-1))/2;
    ll sum=1,last=0,f=1,head=0,tail=0,x1,curr=1;
    parent1[0]=0;
    while(curr)
    {
        x1=curr;
        curr=parent[curr];
    }
    curr=1;
    for(auto i:tree[0])
    {
        last+=i.se*sum;
        sum+=i.se;
        if(i.fi!=x1) curr+=i.se;
    }
    ans[0]-=last;
    for(ll i=1;i<n;i++)
    {
        ll x=find(i);
        if(x==i)
        {
            ans[i]=0;
            continue;
        }
        if(x==head)
        {
            ll sum1=1,sum2=1;
            for(auto j:tree[i])
            {
                if(j.fi!=parent[i]) sum1+=j.se;
            }
            if(tail){
            for(auto j:tree[tail])
            {
                if(j.fi!=parent[tail]) sum2+=j.se;
            }}
            else sum2=curr;
            ans[i]=last-(sum1*sum2);
            last=sum1*sum2;
            head=i;
        }
        else if(x==tail)
        {
            ll sum1=1,sum2=1;
            for(auto j:tree[i])
            {
                if(j.fi!=parent[i]) sum1+=j.se;
            }
            if(head){
            for(auto j:tree[head])
            {
                if(j.fi!=parent[head]) sum2+=j.se;
            }}
            else sum2=curr;
            ans[i]=last-(sum1*sum2);
            last=sum1*sum2;
            tail=i;
        }
        else
        {
           f=0;
           ans[i]=last;
           for(ll j=i+1;j<=n;j++) ans[j]=0;
           break;
        }
    }
    if(f) ans[n]=1;
    for(ll i=0;i<=n;i++) cout<<ans[i]<<" ";
    cout<<endl;
}

int main() {
    ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	ll t=1;
	cin>>t;
	while(t--)
	{
	   solve();
	}
}
```