#include "bits/stdc++.h"
#define rep(i, a, n) for (auto i = a; i <= (n); ++i)
#define revrep(i, a, n) for (auto i = n; i >= (a); --i)
#define all(a) a.begin(), a.end()
#define sz(a) (int)(a).size()
template<class T> inline bool chmax(T &a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T &a, T b) { if (b < a) { a = b; return 1; } return 0; }
using namespace std;

template<class A, class B> string to_string(const pair<A, B> &p);
string to_string(const string &s) { return '"' + s + '"'; }
string to_string(const char *s) { return to_string((string) s); }
string to_string(char c) { return "'" + string(1, c) + "'"; }
string to_string(bool x) { return x ? "true" : "false"; }
template<class A> string to_string(const A &v) {
	bool first = 1;
	string res = "{";
	for (const auto &x: v) {
		if (!first) res += ", ";
		first = 0;
		res += to_string(x);
	}
	res += "}";
	return res;
}
template<class A, class B> string to_string(const pair<A, B> &p) {
	return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}

void debug_out() { cerr << endl; }
template<class H, class... T> void debug_out(const H& h, const T&... t) {
	cerr << " " << to_string(h);
	debug_out(t...);
}
#ifndef ONLINE_JUDGE
	#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
	#define debug(...) if(0) puts("No effect.")
#endif

using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
using vvi = vector<vi>;

/**
 * Author: Yuhao Yao
 * Date: 22-10-25
 * Description: Heavy Light Decomposition for a rooted tree $T$. The root is set as $0$ by default. It can be modified easily for forest.
 *  $g$ should be the adjacent list of the tree $T$.
 *  $chainApply(u, v, func, val)$ and $chainAsk(u, v, func)$ are used for apply / query on the simple path from $u$ to $v$ on tree $T$. $func$ is the function you want to use to apply / query on a interval. (Say rangeApply / rangeAsk of Segment tree.)
 * Time: O(|T|) for building. O(\log |T|) for lca. O(\log |T| \cdot A) for chainApply / chainAsk, where $A$ is the running time of $func$ in chainApply / chainAsk.
 * Status: tested on https://codeforces.com/contest/487/problem/E.
 */

struct HLD {
	int n; /// start-hash
	vi fa, hson, dfn, dep, top;
	HLD(vector<vi> &g, int rt = 0): n(sz(g)), fa(n, -1), hson(n, -1), dfn(n), dep(n, 0), top(n) {
		vi siz(n);
		auto dfs = [&](auto &dfs, int now) -> void {
			siz[now] = 1;
			int mx = 0;
			for (auto v: g[now]) if (v != fa[now]) {
				dep[v] = dep[now] + 1;
				fa[v] = now;
				dfs(dfs, v);
				siz[now] += siz[v];
				if (mx < siz[v]) {
					mx = siz[v];
					hson[now] = v;
				}
			}
		};
		dfs(dfs, rt);

		int cnt = 0;
		auto getdfn = [&](auto &dfs, int now, int sp) {
			top[now] = sp;
			dfn[now] = cnt++;
			if (hson[now] == -1) return;
			dfs(dfs, hson[now], sp);
			for (auto v: g[now]) {
				if(v != hson[now] && v != fa[now]) dfs(dfs, v, v);
			}
		};
		getdfn(getdfn, rt, rt);
	} /// end-hash

	int lca(int u, int v) { /// start-hash
		while (top[u] != top[v]) {
			if (dep[top[u]] < dep[top[v]]) swap(u, v);
			u = fa[top[u]];
		}
		if (dep[u] < dep[v]) return u;
		else return v;
	} /// end-hash

	template<class... T> /// start-hash
	void chainApply(int u, int v, const function<void(int, int, T...)> &func, const T&... val) {
		int f1 = top[u], f2 = top[v];
		while (f1 != f2) {
			if (dep[f1] < dep[f2]) swap(f1, f2), swap(u, v);
			func(dfn[f1], dfn[u], val...);
			u = fa[f1]; f1 = top[u];
		}
		if (dep[u] < dep[v]) swap(u, v);
		func(dfn[v], dfn[u], val...); // change here if you want the info on edges.
	} /// end-hash

	template<class T> /// start-hash
	T chainAsk(int u, int v, const function<T(int, int)> &func) {
		int f1 = top[u], f2 = top[v];
		T ans{};
		while (f1 != f2) {
			if (dep[f1] < dep[f2]) swap(f1, f2), swap(u, v);
			ans = ans + func(dfn[f1], dfn[u]);
			u = fa[f1]; f1 = top[u];
		}
		if (dep[u] < dep[v]) swap(u, v);
		ans = ans + func(dfn[v], dfn[u]); // change here if you want the info on edges.
		return ans;
	} /// end-hash
};

/**
 * Author: Yuhao Yao
 * Date: 22-09-27
 * Description: Fenwick Tree.
 */
template<class T> struct BIT {
	int n;
	vector<T> a;

	BIT(int n): n(n), a(n + 1, 0) {}

	void Add(int i, T x) { 
		for (++i; i <= n; i += i & -i) a[i] += x;
	}

	T Ask(int i) {
		T ans{};
		for (++i; i; i -= i & -i) ans += a[i];
		return ans;
	}

	T rangeAsk(int l, int r) { return Ask(r) - Ask(l - 1); }

	// assuming prefix sums are non-decreasing, finds the first pos such that ask(pos) >= x.
	int lower_bound(T x) {
		assert(n > 0);
		int pos = 0;
		for (int h = 1 << __lg(n); h; h >>= 1) {
			if ((pos | h) <= n && a[pos | h] < x) {
				pos |= h;
				x -= a[pos];
			}
		}
		return pos;
	}
};

int main() {
	ios::sync_with_stdio(0); cin.tie(0);
	int n; cin >> n;
	vvi g(n);
	rep(i, 1, n - 1) {
		int x, y; cin >> x >> y;
		x--; y--;
		g[x].push_back(y);
		g[y].push_back(x);
	}
	HLD hld(g);
	auto fa = hld.fa;
	auto tin = hld.dfn;
	vi tout(n);
	auto getdfn = [&](auto &dfs, int now, int fa) -> void {
		tout[now] = tin[now];
		for (auto v: g[now]) if (v != fa) {
			dfs(dfs, v, now);
			chmax(tout[now], tout[v]);
		}
	};
	getdfn(getdfn, 0, -1);
	vector<pii> pairs;
	rep(x, 1, n) rep(y, x, n) {
		if (1ll * x * y > n) break;
		pairs.emplace_back(x, y);
	}
	sort(all(pairs), [](pii a, pii b) { return 1ll * a.first * a.second < 1ll * b.first * b.second; });
	BIT<ll> bit(n), bit2(n);
	auto ptr = pairs.begin();
	vector<ll> tag(n), tag1(n), tag2(n);
	rep(z, 1, n) {
		int now = z - 1;
		while (ptr != pairs.end() && 1ll * ptr->first * ptr->second < z) {
			auto [x, y] = *ptr;
			ptr++;
			int u = x - 1;
			int v = y - 1;
			int lca = hld.lca(u, v);
			if (lca > 1ll * x * y) {
				tag2[lca]++;
			}
			bit.Add(tin[u], 1);
			bit.Add(tin[v], 1);
			bit.Add(tin[lca], -1);
			if (fa[lca] != -1) bit.Add(fa[lca], -1);
			bit2.Add(tin[u], 1);
			bit2.Add(tin[v], 1);
			bit2.Add(tin[lca], -2);
		}
		tag[now] = bit.rangeAsk(tin[now], tout[now]);
		for (auto v: g[now]) if (v != fa[now]) {
			tag1[v] += tag[now] - bit2.rangeAsk(tin[v], tout[v]);
		}
	}
	ll tot = accumulate(all(tag2), 0ll);
	auto dfs = [&](auto &dfs, int now, int fa) -> void {
		for (auto v: g[now]) if (v != fa) {
			tag1[v] += tag1[now];
			tag2[v] += tag2[now];
			dfs(dfs, v, now);
		}
	};
	dfs(dfs, 0, -1);
	rep(now, 0, n - 1) {
		ll ans = 1ll * n * (n + 1) / 2 - (tag[now] + tag1[now] + tot - tag2[now]);
		printf("%lld\n", ans);
	}
	return 0;
}