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#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl "\n"
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const int N = 200010, M = N << 1;
int n;
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// 链式前向星
int e[M], h[N], idx, w[M], ne[M];
void add(int a, int b, int c = 0) {
e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
}
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int sz[N], f[N], g[N], ans;
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void dfs1(int u, int fa) {
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sz[u] = 1;
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for (int i = h[u]; ~i; i = ne[i]) {
int v = e[i];
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if (v == fa) continue;
dfs1(v, u);
sz[u] += sz[v];
f[u] += f[v];
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}
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f[u] += sz[u];
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}
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void dfs2(int u, int fa) {
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for (int i = h[u]; ~i; i = ne[i]) {
int v = e[i];
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if (v == fa) continue;
g[v] = n - sz[v] + g[u] - sz[v];
dfs2(v, u);
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}
}
signed main() {
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// 初始化链式前向星
memset(h, -1, sizeof h);
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cin >> n;
for (int i = 1; i < n; i++) {
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int a, b;
cin >> a >> b;
add(a, b), add(b, a);
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}
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// 第一次dfs,以子孙节点信息更新父节点的统计信息统计信息包括以u为根的子树中节点数个sz[u],每个节点可以获取到的权值f[u]
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dfs1(1, 0);
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// f[i]:以1为根时的, 以i为根的子树可以获得的最大权值
// g[i]:不管以谁为根,以i为根的子树可以获得的最大权值,也就是最终的结果存储数组
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g[1] = f[1]; // g[1]
// 第二次dfs,换根
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dfs2(1, 0);
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// 遍历一遍历,找出到底以谁为根可以获取到权值的最大值,最大值是多少
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int ans = 0;
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for (int i = 1; i <= n; i++) ans = max(ans, g[i]);
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// 输出答案
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cout << ans << endl;
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}