Problem: Consider a rooted tree. A rooted tree has one special vertex called the root. All edges are directed from the root. Vertex u is called a child of vertex v and vertex v is called a parent of vertex u if there exists a directed edge from v to u. A vertex is called a leaf if it doesn't have children and has a parent.

Let's call a rooted tree a spruce if its every non-leaf vertex has at least 3 leaf children. You are given a rooted tree, check whether it's a spruce.

The definition of a rooted tree can be found [here](https://goo.gl/1dqvzz).

Input

The first line contains one integer n — the number of vertices in the tree (3 ≤ n ≤ 1 000). Each of the next n - 1 lines contains one integer pi (1 ≤ i ≤ n - 1) — the index of the parent of the i + 1-th vertex (1 ≤ pi ≤ i).

Vertex 1 is the root. It's guaranteed that the root has at least 2 children.

Output

Print "Yes" if the tree is a spruce and "No" otherwise.

Examples

Input

4
1
1
1


Output

Yes


Input

7
1
1
1
2
2
2


Output

No


Input

8
1
1
1
1
3
3
3


Output

Yes

Note

The first example:

<image>

The second example:

<image>

It is not a spruce, because the non-leaf vertex 1 has only 2 leaf children.

The third example:

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline int gcd(int a, int b) { return (b == 0 ? a : gcd(b, a % b)); }
inline int lcm(int a, int b) { return (a * (b / gcd(a, b))); }
inline long long int bigmod(long long int a, long long int b, long long int m) {
  if (b == 0) return 1 % m;
  long long int x = bigmod(a, b / 2, m);
  x = (x * x) % m;
  if (b % 2 == 1) {
    x = (x * a) % m;
  }
  return x;
}
inline long long int phi(long long int n) {
  long long int ret = n;
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) n /= i;
    }
    ret -= ret / i;
  }
  if (n > 1) {
    ret -= ret / n;
  }
  return ret;
}
long long int missing_number(long long int A[], long long int n) {
  long long int x1 = A[0], x2 = 1;
  for (int i = 1; i < n; i++) x1 ^= A[i];
  for (int i = 2; i <= n + 1; i++) x2 ^= i;
  return (x1 ^ x2);
}
bool isFibo(long long int n) {
  long long int A = (5 * n * n + 4);
  long long int B = (5 * n * n - 4);
  long long int C = sqrt(A);
  long long int D = sqrt(B);
  return (A == (C * C) || B == (D * D));
}
long long int classini(long long int n) { return (n & 1) ? -1 : 1; }
long long int factorial(int n) {
  return (n == 1 || n == 0) ? 1 : factorial(n - 1) * n;
}
long long int catanal(long long int n) {
  return (factorial(2 * n) / ((factorial(n + 1) * factorial(n))));
}
string orientation_3_ordered_point(vector<pair<int, int> > V) {
  long long int value =
      (V[1].second - V[0].second) * (V[2].first - V[1].first) -
      (V[1].first * V[0].first) * (V[2].second - V[1].second);
  string res;
  if (value == 0) res = "colinear";
  value == 1 ? res = "clockwise" : res = "counterclockwise";
  return res;
}
int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
int fx[] = {+0, +0, +1, -1, -1, +1, -1, +1};
int fy[] = {-1, +1, +0, +0, +1, +1, -1, -1};
string make_string(long long int num) {
  stringstream ss;
  ss << num;
  string s;
  ss >> s;
  return s;
}
long long int make_num(string s) {
  stringstream ss;
  long long int tmp;
  ss << s;
  ss >> tmp;
  return tmp;
}
int SET(int mask, int pos) { return (mask | (1 << pos)); }
int check(int mask, int pos) { return (mask & (1 << pos)); }
vector<int> G[10101];
int add[101010];
int main() {
  int n;
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int v;
    cin >> v;
    G[v].push_back(i);
  }
  for (int i = 1; i <= n; i++)
    if (G[i].empty()) add[i] = 1;
  for (int i = 1; i <= n; i++) {
    if (G[i].empty()) continue;
    int sm = 0;
    for (int j = 0; j < G[i].size(); j++) {
      int v = G[i][j];
      sm += add[v];
    }
    if (sm < 3) {
      puts("No");
      return 0;
    }
  }
  puts("Yes");
  return 0;
}