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;
template <class T>
T sqr(T x) {
  return x * x;
}
const int maxn = 300000;
int vis[maxn], edges[maxn];
int n, m, x, y;
vector<int> g[maxn];
void dfs(int k, int last = -1) {
  vis[k] = 1;
  int v;
  for (int i = 0; i < g[k].size(); i++)
    if ((v = g[k][i]) != last) {
      if (g[v].size() > 2) continue;
      dfs(v, k);
    }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; i++) {
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i] && g[i].size() == 1) dfs(i);
  for (int u = 1; u <= n; u++) {
    for (int i = 0; i < g[u].size(); i++)
      if (vis[g[u][i]]) edges[u] = min(edges[u] + 1, 2);
  }
  for (int u = 1; u <= n; u++) {
    int cnt = 0;
    for (int i = 0; i < g[u].size(); i++) {
      int v = g[u][i];
      if (vis[v]) continue;
      if (g[v].size() > edges[v] + 1) cnt++;
    }
    if (cnt > 2) {
      puts("No");
      return 0;
    }
  }
  puts("Yes");
}
