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. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

# define pb push_back

typedef pair<int64_t, int64_t> pii;
typedef vector<int64_t> vi;
typedef vector<pii> vii;

void pre(){
    
}

bool multi = true;
int timer = 0;

void dfs(
    int p, int u, vector<vector<int> > &adj, vector<int> &tin, 
    vector<int> &tout, vector<int> &count, vector<int> &depth){
    count[u] = 1;
    tin[u] = timer++;
    if (u!=0) depth[u] = depth[p] + 1;
    for(auto v: adj[u]){
        if (v==p) continue;
        dfs(u, v, adj, tin, tout, count, depth);
        count[u]+=count[v];
    }
    tout[u] = timer++;
}

bool is_anc(int a, int u, vector<int> &tin, vector<int> &tout){
    return tin[a]<=tin[u] && tout[a]>=tout[u];
}

int64_t nc2(int64_t n){
    return (n*(n-1))/2;
}

void solve(){
    int n; cin>>n;
    vector<vector<int> > adj(n+1, vector<int> ());
    for(int i=0; i < n-1; i++){
        int x, y; cin>>x>>y;
        adj[x].push_back(y);
        adj[y].push_back(x);
    }
    vector<int> tin(n), tout(n), count(n), depth(n);
    vector<int64_t> res(n+1);
    dfs(-1, 0, adj, tin, tout, count, depth);

    int l_end = 1, r_end = 0;
    int bad = 0;

    vector<int64_t> v = {1};
    int ancl = -1;
    // subtract count other than containing 1
    for(auto e: adj[0]){
        if (is_anc(e, 1, tin, tout)){
            ancl = e;
            count[0] -= count[e];
            v.push_back(count[e] - count[1]);
        }else{
            v.push_back(count[e]);
        }
        res[0]+= nc2(count[e]);
    }

    int64_t sm1 = 0;
    int64_t sm2 = 0;
    for(auto e: v) sm1+=e, sm2+=e*e;
    res[1] = (sm1*sm1 -  sm2) / 2;
    for(int i=2; i < n; i++){
        bool found = false;
        if (is_anc(l_end, i, tin, tout) || is_anc(i, l_end, tin, tout)){
            found = true;
            if (depth[i] > depth[l_end]){ // i is below l_end
                int64_t n1 = count[l_end] - count[i];
                int64_t n2 = count[r_end];
                res[i] = n1*n2;
                l_end = i;
            }
        }
        if ((not found) and (is_anc(r_end, i, tin, tout) || is_anc(i, r_end, tin, tout))){
            found = false;
            for(auto e: adj[0]) if (e!=ancl && is_anc(e, i, tin, tout)) found=true;

            if (found && depth[i] > depth[r_end]){ // i is below l_end
                int64_t n1 = count[r_end] - count[i];
                int64_t n2 = count[l_end];
                res[i] = n1*n2;
                r_end = i;
            }
        }
        if (not found) {
            bad = true;
            res[i] = count[l_end] * 1ll * count[r_end];
        }
        if (bad){
            break;
        }
    }
    if(not bad) res[n] = 1;
    for(auto e: res) cout<<e<<' ';
    cout<<endl;
}

int32_t main(){

#ifdef USE_FILE_IO
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif

#ifndef USE_FILE_IO
    ios_base::sync_with_stdio(0); cin.tie();
#endif
    pre();
    int t = 1; 
    if (multi) cin>>t;
    while(t--) solve();
}