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 = 1e+5 + 1000;
const int INF = numeric_limits<int>::max();
const long long LL_INF = numeric_limits<long long>::max();
struct Edge {
  int to, next;
  Edge() {}
  Edge(int _to, int _next) : to(_to), next(_next) {}
} e[MAXN << 2];
int head[MAXN], tot;
void init() {
  memset(head, -1, sizeof(head));
  tot = 0;
}
void AddEdge(int u, int v) {
  e[tot] = Edge(v, head[u]);
  head[u] = tot++;
  e[tot] = Edge(u, head[v]);
  head[v] = tot++;
}
int du[MAXN], vis[MAXN], res[MAXN];
void dfs(int u) {
  vis[u] = 1;
  for (int i = head[u]; ~i; i = e[i].next) {
    if (du[e[i].to] <= 2 && !vis[e[i].to]) dfs(e[i].to);
  }
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    init();
    memset(du, 0, sizeof(du));
    memset(vis, 0, sizeof(vis));
    memset(res, 0, sizeof(res));
    for (int i = 0, x, y; i < n - 1; i++) {
      scanf("%d%d", &x, &y);
      AddEdge(x, y);
      ++du[x];
      ++du[y];
    }
    for (int i = 1; i <= n; i++)
      if (du[i] == 1) dfs(i);
    int flag = 0;
    for (int u = 1; u <= n; u++) {
      if (!vis[u]) {
        for (int i = head[u]; ~i; i = e[i].next) {
          int v = e[i].to;
          if (vis[v]) res[u] = min(res[u] + 1, 2);
        }
      }
    }
    for (int u = 1; u <= n; u++) {
      if (!vis[u]) {
        int cnt = 0;
        for (int i = head[u]; ~i; i = e[i].next) {
          int v = e[i].to;
          if (!vis[v] && du[v] - res[v] > 1) ++cnt;
        }
        if (cnt > 2) {
          flag = 1;
          break;
        }
      }
    }
    if (flag)
      puts("No");
    else
      puts("Yes");
  }
  return 0;
}
