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;
const int maxn = 100000 + 5;
vector<int> G[maxn];
int leg[maxn];
bool del[maxn];
void dfs(int cur, int per = -1) {
  if (G[cur].size() <= 2) {
    del[cur] = true;
    int siz = G[cur].size();
    for (int i = 0; i < siz; i++) {
      int b = G[cur][i];
      if (b == per) continue;
      dfs(b, cur);
    }
  }
}
int main() {
  int n;
  while (scanf("%d", &n) == 1) {
    memset(del, false, sizeof(del));
    memset(leg, 0, sizeof(leg));
    for (int i = 1; i <= n; i++) G[i].clear();
    for (int i = 1; i < n; i++) {
      int a, b;
      scanf("%d%d", &a, &b);
      G[a].push_back(b);
      G[b].push_back(a);
    }
    for (int i = 1; i <= n; i++)
      if (G[i].size() == 1) {
        dfs(i);
      }
    for (int a = 1; a <= n; a++) {
      int siz = G[a].size();
      for (int j = 0; j < siz; j++) {
        int b = G[a][j];
        if (del[b]) {
          leg[a] = min(leg[a] + 1, 2);
        }
      }
    }
    bool falg = true;
    for (int a = 1; a <= n; a++)
      if (del[a] == false) {
        int cnt = 0;
        for (int j = 0; j < G[a].size(); j++) {
          int b = G[a][j];
          if (del[b] == false && G[b].size() - leg[b] > 1) cnt++;
        }
        if (cnt > 2) {
          falg = false;
          break;
        }
      }
    if (falg)
      puts("YES");
    else
      puts("NO");
  }
  return 0;
}
