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 = 0;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = 1;
  for (; ch >= '0' && ch <= '9'; ch = getchar())
    x = (x << 1) + (x << 3) + ch - 48;
  return f ? -x : x;
}
int n, in[100005], cnt, head[100005];
struct edge {
  int next, to;
} e[100005 << 1];
inline void insert(int u, int v) {
  e[++cnt] = (edge){head[u], v};
  head[u] = cnt;
  in[u]++;
  e[++cnt] = (edge){head[v], u};
  head[v] = cnt;
  in[v]++;
}
bool lef[100005];
int Cnt[100005];
inline void dfs(int x, int fa) {
  lef[x] = 1;
  for (int i = head[x]; i; i = e[i].next)
    if (e[i].to != fa && in[e[i].to] <= 2) dfs(e[i].to, x);
}
int main() {
  n = read();
  for (int i = 1; i <= n - 1; i++) {
    int u = read(), v = read();
    insert(u, v);
  }
  for (int i = 1; i <= n; i++)
    if (in[i] == 1) dfs(i, 0);
  for (int x = 1; x <= n; x++)
    for (int i = head[x]; i; i = e[i].next)
      if (lef[e[i].to]) Cnt[x]++;
  for (int x = 1; x <= n; x++)
    if (!lef[x]) {
      int sum = 0;
      for (int i = head[x]; i; i = e[i].next)
        if (!lef[e[i].to] && in[e[i].to] - min(Cnt[e[i].to], 2) >= 2) sum++;
      if (sum >= 3) return puts("No") & 0;
    }
  puts("Yes");
  return 0;
}
