#include <bits/stdc++.h>
#define FOR(i, l, r) for(int i = l; i <= r; ++i)
#define FOD(i, r, l) for(int i = r; i >= l; --i)
#define debug(x) cout << #x << " = " << x << '\n';

using namespace std;

const int maxn = 2*1e5 + 11;

int t, n, l, r;
long long P, ans;
vector < int > g[maxn];
int par[maxn], avail[maxn], sizeChild[maxn];

void dfs(int u) {
    sizeChild[u] = 1;
    for(auto &v : g[u]) {
        if(par[u] == v) continue;
        par[v] = u;
        dfs(v);
        sizeChild[u] += sizeChild[v];
        continue;
    }
}

int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while(t--) {
        cin >> n;
        FOR(i, 0, n - 1) {
            g[i].clear();
            par[i] = -1;
            avail[i] = 0;
            sizeChild[i] = 0;
        }
        ans = 0;
        P = 1ll*n*(n - 1)/2;
        l = r = 0;
        avail[0] = 1;

        FOR(i, 1, n - 1) {
            int u, v;
            cin >> u >> v;
            g[u].push_back(v);
            g[v].push_back(u);
        }

        dfs(0);
        FOR(i, 0, n - 1) {
            if(i == 0) {
                ans = 0;
                for(auto &v : g[i])
                    ans += 1ll * sizeChild[v] * (sizeChild[v] - 1)/2;
                P -= ans;
                cout << ans << " ";
            }
            else if(P == 0 && ans == 0) cout << 0 << " ";
            else {
                if(avail[i]) {
                    cout << 0 << " ";
                    continue;
                }
                int u = i;
                while(true) {
                    avail[u] = 1;
                    if(par[u] == 0) sizeChild[0] -= sizeChild[u];
                    u = par[u];
                    if(avail[u]) {
                        if(u != l && u != r) {
                            cout << P << " ";
                            P = ans = 0;
                            break;
                        }
                        if(u == l)
                            l = i;
                        else r = i;
                        ans = P - 1ll * sizeChild[l] * sizeChild[r];
                        cout << ans << " ";
                        P -= ans;
                        break;
                    }
                }
            }
        }
        cout << P << '\n';
    }
    return 0;
}

### Prompt

Develop a solution in Cpp 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 FOR(i, l, r) for(int i = l; i <= r; ++i)
#define FOD(i, r, l) for(int i = r; i >= l; --i)
#define debug(x) cout << #x << " = " << x << '\n';

using namespace std;

const int maxn = 2*1e5 + 11;

int t, n, l, r;
long long P, ans;
vector < int > g[maxn];
int par[maxn], avail[maxn], sizeChild[maxn];

void dfs(int u) {
    sizeChild[u] = 1;
    for(auto &v : g[u]) {
        if(par[u] == v) continue;
        par[v] = u;
        dfs(v);
        sizeChild[u] += sizeChild[v];
        continue;
    }
}

int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while(t--) {
        cin >> n;
        FOR(i, 0, n - 1) {
            g[i].clear();
            par[i] = -1;
            avail[i] = 0;
            sizeChild[i] = 0;
        }
        ans = 0;
        P = 1ll*n*(n - 1)/2;
        l = r = 0;
        avail[0] = 1;

        FOR(i, 1, n - 1) {
            int u, v;
            cin >> u >> v;
            g[u].push_back(v);
            g[v].push_back(u);
        }

        dfs(0);
        FOR(i, 0, n - 1) {
            if(i == 0) {
                ans = 0;
                for(auto &v : g[i])
                    ans += 1ll * sizeChild[v] * (sizeChild[v] - 1)/2;
                P -= ans;
                cout << ans << " ";
            }
            else if(P == 0 && ans == 0) cout << 0 << " ";
            else {
                if(avail[i]) {
                    cout << 0 << " ";
                    continue;
                }
                int u = i;
                while(true) {
                    avail[u] = 1;
                    if(par[u] == 0) sizeChild[0] -= sizeChild[u];
                    u = par[u];
                    if(avail[u]) {
                        if(u != l && u != r) {
                            cout << P << " ";
                            P = ans = 0;
                            break;
                        }
                        if(u == l)
                            l = i;
                        else r = i;
                        ans = P - 1ll * sizeChild[l] * sizeChild[r];
                        cout << ans << " ";
                        P -= ans;
                        break;
                    }
                }
            }
        }
        cout << P << '\n';
    }
    return 0;
}

```