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 int long long
//#define DB
#ifdef DB
#define el cerr << "\n";
    #define db(...) cerr << " [" << #__VA_ARGS__ << " : " << __VA_ARGS__ << "] ";
#else
#define el
#define db(...)
#endif // DB
const int N = 1e6 + 5;
int n;
vector<int> adj[N];
int sz[N];
int par[N];

void dfs(int u = 0, int p = -1) {
    par[u] = p;
    sz[u] = 1;
    for (int v : adj[u]) {
        if (v == p) continue;
        dfs(v, u);
        sz[u] += sz[v];
    }
}

bool visited[N];
int P[N];

void solve() {
    cin >> n;
    for (int i = 0; i < n; ++i) {
        adj[i].clear();
        visited[i] = false;
        sz[i] = P[i] = 0;
        par[i] = -1;
    }
    P[n] = 0;
    int u, v;
    for (int i = 1; i < n; ++i) {
        cin >> u >> v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    dfs();
    int sum = 1;
    for (int v : adj[0]) {
        P[0] += sz[v] * sum;
        sum += sz[v];
    }
    int L = 0, R = 0, j, k;
    int szL, szR;
    visited[0] = true;
    for (int i = 1; i < n; ++i) {
//        db(i) el;
        if (visited[i]) {
            P[i] = P[i - 1];
            continue;
        }
        j = i, k = i;
        while (!visited[j]) {
            visited[j] = true;
            k = j;
            j = par[j];
        }
        if (j != L and j != R) {
            break;
        }
        if (j == L) {
            if (L == 0) {
                szR = n - sz[k];
            }
            L = i;
            szL = sz[L];
        } else {
            R = i;
            szR = sz[R];
        }
//        db(i) db(L) db(R) db(szL) db(szR) el;
        P[i] = szL * szR;
    }
    int cur = n * (n - 1) / 2;
    for (int i = 0; i <= n; ++i) {
//        cerr << P[i] << " ";
        cout << cur - P[i] << " ";
        cur = P[i];
    }
//    el;
    cout << "\n";
}
#undef int
int main() {
    // freopen("C:\\Users\\T\\CLionProjects\\cp-chemthan\\codeforces\\contest\\div2\\round721\\D.in", "r", stdin);
    ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}