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 i64 = long long;
using u64 = unsigned long long;
using u32 = unsigned;
int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    int t;
    std::cin >> t;
    while (t--) {
        int n;
        std::cin >> n;
        std::vector<std::vector<int>> e(n);
        for (int i = 0; i < n - 1; i++) {
            int u, v;
            std::cin >> u >> v;
            e[u].push_back(v);
            e[v].push_back(u);
        }
        std::vector<i64> ans(n + 1);
        int timeStamp = 0;
        std::vector<int> in(n), out(n), siz(n);
        std::function<void(int, int)> dfs = [&](int u, int p) {
            in[u] = timeStamp++;
            for (auto v : e[u]) {
                if (v == p) {
                    continue;
                }
                dfs(v, u);
            }
            out[u] = timeStamp;
            siz[u] = out[u] - in[u];
        };
        dfs(0, -1);
        ans[0] = ans[1] = i64(n) * (n - 1) / 2;
        int l = -1;
        for (auto v : e[0]) {
            ans[1] -= i64(siz[v]) * (siz[v] - 1) / 2;
            if (in[v] <= in[1] && in[1] < out[v]) {
                l = v;
            }
        }
        int u = 1, v = -1;
        for (int i = 1; i < n; i++) {
            bool ok = false;
            if (in[i] <= in[u] && in[u] < out[i]) {
                ok = true;
            }
            if (v != -1 && in[i] <= in[v] && in[v] < out[i]) {
                ok = true;
            }
            if (!ok) {
                if (in[u] <= in[i] && in[i] < out[u]) {
                    u = i;
                } else if ((v == -1 && !(in[l] <= in[i] && in[i] < out[l])) || (in[v] <= in[i] && in[i] < out[v])) {
                    v = i;
                } else {
                    break;
                }
            }
            if (v == -1) {
                ans[i + 1] = i64(siz[u]) * (n - siz[l]);
            } else {
                ans[i + 1] = i64(siz[u]) * siz[v];
            }
        }
        for (int i = 0; i < n; i++) {
            ans[i] -= ans[i + 1];
        }
        for (int i = 0; i <= n; i++) {
            std::cout << ans[i] << " \n"[i == n];
        }
    }
    return 0;
}