Example

Input

4
1 1 2


Output

6
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=a;i<b;i++)
typedef long long ll;

int N;
vector< pair<int, int> > E[101010];

class LCA
{
public:
	int NV;
	int logNV;
	vector<int> D;
	vector<long long> L;
	vector<vector<int> > P;

	LCA(int N)
	{
		NV = N;
		logNV = 0;
		while (NV > (1LL << logNV)) logNV++;
		D = vector<int>(NV);
		L = vector<long long>(NV);
		P = vector<vector<int> >(logNV, vector<int>(NV));
	}

	void init()
	{
		dfs(0, -1, 0, 0);
		build();
	}

	void dfs(int v, int par, int d, long long l)
	{
		D[v] = d;
		P[0][v] = par;
		L[v] = l;

		for (int i = 0; i < (int)E[v].size(); i++) {
			int w = E[v][i].first;
			int lc = E[v][i].second;
			if (w == par) continue;
			dfs(w, v, d + 1, lc + l);
		}
	}

	void build() {
		for (int k = 0; k + 1 < logNV; k++) {
			for (int v = 0; v < NV; v++) {
				if (P[k][v] < 0) P[k + 1][v] = -1;
				else P[k + 1][v] = P[k][P[k][v]];
			}
		}
	}

	int query(int u, int v) {
		if (D[u] > D[v]) swap(u, v);
		for (int k = 0; k < logNV; k++) {
			if ((D[v] - D[u]) >> k & 1)
				v = P[k][v];
		}
		if (u == v) return u;

		for (int k = logNV - 1; k >= 0; k--) {
			if (P[k][u] != P[k][v]) {
				u = P[k][u];
				v = P[k][v];
			}
		}
		return P[0][u];
	}
};

int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	while (cin >> N)
	{
		if (N == 1)
		{
			cout << 0 << endl;
			continue;
		}
		rep(i, 0, N) E[i].clear();
		rep(i, 1, N)
		{
			int p; cin >> p; p--;
			E[p].push_back( make_pair(i, 1) );
			E[i].push_back( make_pair(p, 1) );
		}

		LCA lca(N);
		lca.init();

		ll ans = 0;

		queue<int> que;
		que.push(0);
		int prev = -1;
		while (!que.empty())
		{
			int p = que.front(); que.pop();
			for (auto _p : E[p])
			{
				int pp = _p.first;
				if (lca.L[p] < lca.L[pp]) que.push(pp);
			}

			if (0 <= prev)
			{
				ans += (ll)(lca.L[p] + lca.L[prev] - 2 * lca.L[lca.query(p, prev)]);
			}
			prev = p;
		}
		cout << ans << endl;
	}
}