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 = 1e5 + 100;
int num[maxn], vis[maxn], n;
int f[maxn];
int g[maxn];
vector<int> G[maxn];
void dp(int u) {
  vis[u] = 1;
  num[u] = 0;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (!vis[v]) dp(v);
    if (G[v].size() - 1 > 2) {
      num[u]++;
    }
  }
}
void dfs_f(int u, int fa) {
  f[u] = 0;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (v != fa) {
      dfs_f(v, u);
      if (f[v] > 0 || G[v].size() - 1 > 1) f[u]++;
    }
  }
}
int dfs(int u, int fa, int flag) {
  int tmp = 0;
  int sonSize = 0;
  int cnt = 0;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (v != fa) {
      sonSize++;
      int ans;
      if (flag != 0) {
        ans = dfs(v, u, 2);
      } else {
        if (G[v].size() - 1 > 2) {
          cnt++;
          ans = dfs(v, u, 0);
        } else if (f[v] > 0) {
          cnt++;
          ans = dfs(v, u, 0);
        } else {
          ans = dfs(v, u, 1);
        }
      }
      if (ans == 0) return 0;
    }
  }
  if (fa == 0) {
    if (cnt > 2) return 0;
    return 1;
  }
  if (cnt > 1) return 0;
  if (flag == 1) {
    if (sonSize <= 2) return 1;
    return 0;
  } else if (flag == 2) {
    if (sonSize <= 1) return 1;
    return 0;
  }
  return 1;
}
int tag = 0, tagrt;
void change(int u, int fa) {
  if (tag) return;
  g[u] = f[u];
  int cnt = 0;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (G[v].size() - 1 > 2 || f[v] >= 1) {
      cnt++;
    }
  }
  if (cnt >= 2) {
    tag = 1;
    tagrt = u;
    return;
  }
  int old_u, old_v;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (v != fa) {
      old_u = f[u];
      old_v = f[v];
      if (G[v].size() - 1 >= 2 || f[v] > 0) {
        f[u]--;
      }
      if (G[u].size() - 1 >= 2 || f[u] > 0) {
        f[v]++;
      }
      change(v, u);
      f[u] = old_u;
      f[v] = old_v;
    }
  }
}
int solve() {
  int key = 1;
  dfs_f(1, 0);
  tag = 0;
  change(1, 0);
  if (tag == 0) return 1;
  dfs_f(tagrt, 0);
  if (dfs(tagrt, 0, 0)) return 1;
  return 0;
}
int main() {
  while (scanf("%d", &n) != EOF) {
    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);
    }
    if (solve())
      printf("Yes\n");
    else
      printf("No\n");
  }
  return 0;
}
