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>
const int inf = 0x3f3f3f3f;
const double eps = 1e-9;
using namespace std;
const int N = 1e5 + 5;
vector<int> edge[N], e[N];
int vis[N], cnt[N], ban[N];
void dfs(int x) {
  cnt[x]--;
  if (edge[x].size() <= 2) {
    ban[x] = vis[x] = 1;
    for (auto &y : edge[x]) {
      if (!vis[y]) {
        dfs(y);
      }
    }
  }
}
bool use[N];
int bfs(int x) {
  queue<int> q;
  q.push(x), use[x] = 1;
  int ans = x;
  while (q.size()) {
    int x = q.front();
    q.pop();
    for (auto &y : edge[x]) {
      if (!use[y] && ban[y] != 1) {
        use[y] = 1;
        q.push(y);
        ans = y;
      }
    }
  }
  return ans;
}
bool DFS(int x, int p = 0) {
  int cnt = 0;
  for (auto &y : edge[x])
    if (y != p) {
      if (vis[y] != 1 && !ban[y]) {
        if (!DFS(y, x)) return 0;
        cnt++;
      }
    }
  return cnt <= 1;
}
int main() {
  ios ::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = (int)(1); i <= (int)(n - 1); ++i) {
    int x, y;
    cin >> x >> y;
    edge[x].push_back(y);
    edge[y].push_back(x);
  }
  for (int i = (int)(1); i <= (int)(n); ++i) cnt[i] = edge[i].size();
  for (int i = (int)(1); i <= (int)(n); ++i)
    if (edge[i].size() == 1) {
      dfs(i);
    }
  for (int i = (int)(1); i <= (int)(n); ++i)
    if (!vis[i] && cnt[i] == 1 && cnt[i] != int(edge[i].size())) {
      ban[i] = 1;
    }
  for (int i = (int)(1); i <= (int)(n); ++i)
    if (edge[i].size() > 3) {
      ban[i] = 0;
    }
  int ans = 1;
  for (int i = (int)(1); i <= (int)(n); ++i)
    if (ban[i] == 0) {
      int start = bfs(i);
      ans &= DFS(start);
      break;
    }
  ans ? puts("Yes") : puts("No");
  return 0;
}
