A Ancient Distance

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
//#include<regex>
#include<cstdio>
#include <iomanip>
#pragma GCC optimize(2)
#define up(i,a,b)  for(int i=a;i<b;i++)
#define dw(i,a,b)  for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
	char ch = getchar(); ll x = 0, f = 1;
	while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
	return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 2e5 + 10;
vector<int>vec[N];
int n;
int st = 0;
int f[N], dfn[N], ed[N], par[N][25];
int rk[N];
struct seg {
	int mx[N << 2];
	int mxx[N << 2];
	void pushup(int root)
	{
		mx[root] = max(mx[root << 1], mx[root << 1 | 1]);
	}
	void build(int l, int r, int root)
	{
		if (l == r) {
			mx[root] = mxx[root] = f[rk[l]]; return;
		}
		int mid = (l + r) >> 1;
		build(lson);
		build(rson);
		pushup(root);
		mxx[root] = mx[root];
	}
	void update(int l, int r, int root, int L, int R)
	{
		if (L <= l && r <= R)
		{
			mx[root] = -1; return;
		}
		int mid = (l + r) >> 1;
		if (L <= mid)update(lson, L, R);
		if (R > mid)update(rson, L, R);
		pushup(root);
	}
	void rebuild(int l, int r, int root, int L, int R)
	{
		mx[root] = mxx[root];
		if (L <= l && r <= R)
		{
		 return;
		}
		int mid = (l + r) >> 1;
		if (L <= mid)rebuild(lson, L, R);
		if (R > mid)rebuild(rson, L, R);
		pushup(root);
	}
	int query(int l, int r, int root)
	{
		if (l == r)return l;
		int mid = (l + r) >> 1;
		if (mx[root] == -1)return -1;
		if (mx[root << 1] < mx[root << 1 | 1])return query(rson);
		else return query(lson);
	}
}T;
void dfs(int u, int fa)
{
	par[u][0] = fa;
	upd(i, 1, 22)
	{
		par[u][i] = par[par[u][i - 1]][i - 1];
	}
	f[u] = f[fa] + 1;
	dfn[u] = ++st;
	rk[st] = u;
	for (auto k : vec[u])
	{
		if (k == fa)continue;
		dfs(k, u);
	}
	ed[u] = st;
}
vector<pir>temp;
vector<int>ans;
int main()
{
	while (~scanf("%d", &n))
	{
		st = 0;
		upd(i, 1, n)vec[i].clear();
		ans.clear();
		int u;
		upd(i, 2, n)
		{
			u = read(); vec[i].push_back(u); vec[u].push_back(i);
		}
		int need = 0;
		dfs(1, 1);
		T.build(1, st, 1);
		ans.resize(n + 1, -1);
		ans[0] = n;
		for (int d = 1; d <= n-1; d++) {
			int cnt = 0;
			temp.clear();
			while (1) {
				int last = T.query(1, st, 1);
				if (last == -1)break;
				last = rk[last];
				cnt++;
				dwd(i, 22, 0)
				{
					if (d >> i & 1)last = par[last][i];
				}
				temp.push_back({ dfn[last],ed[last] });
				T.update(1, st, 1, dfn[last], ed[last]);
			}
			ans[d] = cnt;
			for (auto k : temp)
			{
				T.rebuild(1, st, 1, k.first, k.second);
			}
			if (cnt == 1)break;
		}
		ll sum = 0;
		upd(d, 1, n-1)
		{
			if (ans[d] == -1)break;                     
			sum += 1ll * (ans[d - 1] - ans[d])*d;
			//if (ans[d] == 1)break;
		}
		printf("%lld\n", sum);
	}
	return 0;
}