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;
typedef long long ll;
ll t = 1,n,m,k,q,fl,cur,cc,cnt,wc1,wc2,l,r,a,b,c,u,v,x,y,z,hd[200009],root[200009],sz[200009],ans[200009],pa[200009],mod = 1e9 + 7;
string s;
bool vis[200009];
struct st{ll x,y;}eg[400009];
bool cmp(const st &a,const st &b){return a.x < b.x;}
void addedge(ll u,ll v){eg[++ cc] = (st){v,hd[u]},hd[u] = cc;} 
inline ll read(){
	ll s = 0,w = 1;
	char ch = getchar();
	while (ch > '9' || ch < '0'){ if (ch == '-') w = -1; ch = getchar();}
	while (ch <= '9' && ch >= '0') s = (s << 1) + (s << 3) + (ch ^ 48),ch = getchar();
	return s * w;
}
ll qp(ll x,ll y){
	ll a = 1,b = x;
	while (y){
		if (y & 1) a = a * b % mod;
		b = b * b % mod,y >>= 1;
	}
	return a;
}
void upd(){ while (vis[cur + 1]) cur += 1;}
void dfs(ll u,ll fa,ll rt){
	sz[u] = 1,pa[u] = fa,root[u] = rt;
	for (ll i = hd[u];i;i = eg[i].y){
		ll v = eg[i].x;
		if (v == fa) continue;
		dfs(v,u,rt),sz[u] += sz[v];
	}
}
void work(ll x){
	if (!cnt){
		ll fk = x;
		while (fk != 1) vis[fk] = 1,fk = pa[fk];
		upd(),ans[cur] += (n - sz[root[x]]) * sz[x];
		cnt += 1,wc1 = x;
	}
	else if (cnt == 1){
		if (root[x] == root[wc1]){
			ll fk = x;
			while (fk != 1 && fk != wc1) vis[fk] = 1,fk = pa[fk];
			if (fk == 1){fl = 0; return;}
			upd(),ans[cur] += (n - sz[root[x]]) * sz[x];
			wc1 = x;
		}
		else{
			ll fk = x;
			while (fk != 1) vis[fk] = 1,fk = pa[fk];
			upd(),ans[cur] += sz[wc1] * sz[x];
			cnt += 1,wc2 = x;
		}
	}
	else{
		ll fk = x;
		while (fk != 1 && !(fk == wc1 || fk == wc2)) vis[fk] = 1,fk = pa[fk];
		if (fk == 1){fl = 0; return;}
		upd();
		if (fk == wc1) wc1 = x;
		else wc2 = x;
		ans[cur] += sz[wc1] * sz[wc2];
	}
}
int main(){t = read();
while (t --){
	n = read(),cur = 1,fl = 1,cnt = 0,wc1 = wc2 = 0,cc = 0;
	fill(vis,vis + n + 5,0),fill(ans,ans + n + 5,0),fill(hd,hd + n + 5,0);
	for (ll i = 1;i < n;i += 1) x = read() + 1,y = read() + 1,addedge(x,y),addedge(y,x);
	ans[1] = n * (n - 1) / 2;
	for (ll i = hd[1];i;i = eg[i].y) dfs(eg[i].x,1,eg[i].x),ans[1] -= sz[eg[i].x] * (sz[eg[i].x] - 1) / 2;
	for (ll i = 2;i <= n && fl;i += 1){
		if (cur > i || vis[i]) continue;
		work(i);
	}
	ans[0] = n * (n - 1) / 2 - ans[n + 1];
	for (ll i = n;i >= 1;i -= 1) if (!ans[i]) ans[i] = ans[i + 1];
	for (ll i = 1;i <= n;i += 1) ans[i] -= ans[i + 1],ans[0] -= ans[i];
	for (ll i = 0;i <= n;i += 1) printf("%lld ",ans[i]);
	puts("");
}
	return 0;
}