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 long long read() {
  long long x = 0;
  int ch = getchar(), f = 1;
  while (!isdigit(ch) && (ch != '-') && (ch != EOF)) ch = getchar();
  if (ch == '-') {
    f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int N = 1e5 + 10;
int n, x, y;
bool vis[N];
int tot, first[N], to[N << 1], last[N << 1];
inline void Add(int x, int y) {
  to[++tot] = y, last[tot] = first[x], first[x] = tot;
}
bool flag;
int dp[N];
inline void dfs(int u, int fa) {
  int cnt = 0;
  for (register int i = first[u]; i; i = last[i]) {
    int v = to[i];
    if (v != fa) dfs(v, u), dp[u] += dp[v], ++cnt;
  }
  if (dp[u] == 2 && cnt == 1 || !dp[u]) ++dp[u];
}
inline void Dfs(int u, int fa, int f) {
  int sum = 0;
  if (f > 2) ++sum;
  for (register int i = first[u]; i; i = last[i]) {
    int v = to[i];
    if (v != fa) {
      int tmp = max(1, f + dp[u] - dp[v]);
      if (tmp == 2 && f == 2) tmp = 3;
      Dfs(v, u, tmp);
      if (dp[v] > 2) sum++;
    }
  }
  if (sum > 2) flag = 0;
}
int main() {
  n = read();
  for (register int i = (1); i <= (n - 1); i++)
    x = read(), y = read(), Add(x, y), Add(y, x);
  flag = 1, dfs(1, 0), Dfs(1, 0, 0);
  puts(flag ? "YES" : "NO");
}
