Limak is a little bear who learns to draw. People usually start with houses, fences and flowers but why would bears do it? Limak lives in the forest and he decides to draw a tree.

Recall that tree is a connected graph consisting of n vertices and n - 1 edges.

Limak chose a tree with n vertices. He has infinite strip of paper with two parallel rows of dots. Little bear wants to assign vertices of a tree to some n distinct dots on a paper so that edges would intersect only at their endpoints — drawn tree must be planar. Below you can see one of correct drawings for the first sample test.

<image>

Is it possible for Limak to draw chosen tree?

Input

The first line contains single integer n (1 ≤ n ≤ 105).

Next n - 1 lines contain description of a tree. i-th of them contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi) denoting an edge between vertices ai and bi. It's guaranteed that given description forms a tree.

Output

Print "Yes" (without the quotes) if Limak can draw chosen tree. Otherwise, print "No" (without the quotes).

Examples

Input

8
1 2
1 3
1 6
6 4
6 7
6 5
7 8


Output

Yes


Input

13
1 2
1 3
1 4
2 5
2 6
2 7
3 8
3 9
3 10
4 11
4 12
4 13


Output

No
#include <bits/stdc++.h>
using namespace std;
vector<int> eh[100005];
int son[100005], dp[100005];
int N, S, T;
bool ok;
void dfs(int u, int f) {
  if (dp[u] != -1) return;
  dp[u] = 0;
  int four = 0, three = 0, two = 0, one = 0;
  for (int i = 0, num = eh[u].size(); i < num; i++) {
    int v = eh[u][i];
    if (v == f) continue;
    dfs(v, u);
    if (dp[v] & 8) four++;
    if ((dp[v] & 2) && !(dp[v] & 4)) two++;
    if (dp[v] == 1) {
      if (S + T == 0) S = u, T = v;
      one++;
    }
  }
  if (son[u] <= 1 && four == son[u]) dp[u] |= 8;
  if (son[u] <= 2 && four == son[u]) dp[u] |= 4;
  if (two <= 1) dp[u] |= 2;
  if (two <= 2) dp[u] |= 1;
  if (two > 2 || one > 1) ok = false;
  return;
}
void judgeline(int u, int f) {
  if (eh[u].size() - 1 > 1) {
    ok = false;
    return;
  }
  for (int i = 0, num = eh[u].size(); i < num; i++) {
    int v = eh[u][i];
    if (v == f) continue;
    judgeline(v, u);
  }
}
int main() {
  while (scanf("%d", &N) != EOF) {
    for (int i = 1; i <= N; i++) eh[i].clear();
    memset(son, 0, sizeof(son));
    memset(dp, -1, sizeof(dp));
    int u, v;
    for (int i = 1; i <= N - 1; i++) {
      scanf("%d %d", &u, &v);
      eh[u].push_back(v);
      eh[v].push_back(u);
    }
    son[1] = eh[1].size();
    for (int i = 2; i <= N; i++) son[i] = eh[i].size() - 1;
    ok = true;
    S = 0, T = 0;
    dfs(1, -1);
    if (S + T) {
      int ss = 0;
      for (int i = 0, num = eh[S].size(); i < num; i++) {
        int v = eh[S][i];
        if (v == T) continue;
        ss++;
        if (ss > 2) {
          ok = false;
          break;
        }
        judgeline(v, S);
      }
    }
    if (ok)
      printf("Yes\n");
    else
      printf("No\n");
  }
  return 0;
}
