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;
struct edge {
  int to, next;
} e[100005 * 2];
int head[100005], deg[100005], vis[100005], leaf[100005], cnt[100005];
int n, x, y, num, tot;
void add(int x, int y) {
  e[++tot] = (edge){y, head[x]};
  head[x] = tot;
}
void dfs(int x, int fa) {
  leaf[x] = 1;
  for (int i = head[x]; i; i = e[i].next)
    if (e[i].to != fa && deg[e[i].to] <= 2) dfs(e[i].to, x);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
    deg[x]++;
    deg[y]++;
  }
  for (int i = 1; i <= n; i++)
    if (deg[i] == 1) dfs(i, 0);
  for (int i = 1; i <= n; i++)
    for (int j = head[i]; j; j = e[j].next)
      if (leaf[e[j].to]) cnt[i]++;
  for (int i = 1; i <= n; i++) {
    if (leaf[i]) continue;
    num = 0;
    for (int j = head[i]; j; j = e[j].next)
      if (!leaf[e[j].to] && deg[e[j].to] - min(cnt[e[j].to], 2) >= 2) num++;
    if (num > 2) return puts("No"), 0;
  }
  puts("Yes");
}
