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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int maxn = 1e5 + 5;
struct edge {
  int link, next;
} e[maxn << 1];
int tot, head[maxn], n, du[maxn];
inline void add(int u, int v) {
  e[++tot] = (edge){v, head[u]};
  head[u] = tot;
}
inline void insert(int u, int v) {
  add(u, v);
  add(v, u);
}
inline void init() {
  n = read();
  for (int i = 1; i < n; i++) {
    int u = read(), v = read();
    insert(u, v);
    du[u]++;
    du[v]++;
  }
}
int d[maxn];
bool vis[maxn];
inline void dfs(int u) {
  vis[u] = 1;
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].link;
    if (!vis[v] && du[v] <= 2) {
      dfs(v);
    }
  }
}
inline void solve() {
  for (int i = 1; i <= n; i++) {
    if (du[i] == 1) {
      dfs(i);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      for (int j = head[i]; j; j = e[j].next) {
        int v = e[j].link;
        if (vis[v]) {
          d[i] = min(d[i] + 1, 2);
        }
      }
    }
  }
  bool flag = 1;
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      int cnt = 0;
      for (int j = head[i]; j; j = e[j].next) {
        int v = e[j].link;
        if (!vis[v] && du[v] - d[v] > 1) {
          cnt++;
        }
      }
      if (cnt > 2) {
        flag = false;
        break;
      }
    }
  }
  if (flag)
    puts("Yes");
  else
    puts("No");
}
int main() {
  init();
  solve();
  return 0;
}
