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> e[200514];
int deg[200514], vis[200514];
int n;
int q[200514], qs, qe;
int cnt[200514];
int dp[200514];
int judge3(int w) {
  for (auto t : e[w])
    if (vis[t]) {
      if (dp[t] >= 2 || (dp[t] == 1 && ((int)e[t].size()) != 3)) return 0;
    }
}
int judge4(int w, int all3) {
  int cnt = 0;
  int sum3 = (((int)e[w].size()) == 3);
  int flag = 0;
  for (auto t : e[w]) {
    if (vis[t] == 0) {
      flag = 1;
      continue;
    }
    sum3 += dp[t];
    cnt += (dp[t] >= 2 || (dp[t] == 1 && ((int)e[t].size()) != 3));
    if (cnt >= 3) return 0;
  }
  if (flag) cnt += (all3 - sum3 >= 2);
  return cnt < 3;
}
int sol(int uuu) {
  for (int i = 0; i < (n); i++) e[i].clear();
  for (int i = 0; i < (n - 1); i++) {
    int x, y;
    scanf("%d %d", &(x), &(y));
    x--, y--;
    e[x].push_back(y);
    e[y].push_back(x);
  }
  for (int i = 0; i < (n); i++) deg[i] = ((int)e[i].size());
  qs = qe = 0;
  int all3 = 0;
  for (int i = 0; i < (n); i++) {
    if (deg[i] == 3) all3++;
    if (deg[i] == 1) {
      q[qe++] = i;
      vis[i] = 1;
    }
  }
  int cnt = 0;
  int x = -1, y = -1;
  for (; qs != qe;) {
    int w = q[qs++];
    dp[w] += (((int)e[w].size()) == 3);
    if (((int)e[w].size()) == 3) {
      if (judge4(w, all3) == 0) return 0;
    }
    for (auto t : e[w]) {
      if (vis[t]) continue;
      deg[t]--;
      dp[t] += dp[w];
      if (deg[t] == 1) {
        if (((int)e[t].size()) <= 3) {
          vis[t] = 1;
          q[qe++] = t;
        } else {
          cnt++;
          if (cnt >= 3) return 0;
          if (cnt == 1) x = t;
          if (cnt == 2) y = t;
        }
      }
    }
  }
  for (int i = 0; i < (n); i++)
    if (vis[i] == 0 && ((int)e[i].size()) >= 3 && i != x && i != y &&
        judge3(i) == 0)
      return 0;
  if (cnt >= 1 && judge4(x, all3) == 0) return 0;
  if (cnt >= 2 && judge4(y, all3) == 0) return 0;
  return 1;
}
int main() {
  while (scanf("%d", &(n)) != EOF) {
    int res = sol(0);
    puts(res ? "Yes" : "No");
  }
  return 0;
}
