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 k = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    k = k * 10 + ch - '0';
    ch = getchar();
  }
  return k * f;
}
int nedge = 0, p[200010], nex[200010], head[200010];
int n, d[200010], ll[200010];
bool lian[200010];
inline void addedge(int a, int b) {
  p[++nedge] = b;
  nex[nedge] = head[a];
  head[a] = nedge;
}
inline void dfs(int x, int fa) {
  lian[x] = 1;
  for (int k = head[x]; k; k = nex[k])
    if (p[k] != fa && d[p[k]] <= 2) dfs(p[k], x);
}
int main() {
  n = read();
  for (int i = 1; i < n; i++) {
    int x = read(), y = read();
    addedge(x, y);
    addedge(y, x);
    d[x]++;
    d[y]++;
  }
  for (int i = 1; i <= n; i++)
    if (d[i] == 1) dfs(i, 0);
  for (int i = 1; i <= n; i++) {
    for (int k = head[i]; k; k = nex[k])
      if (lian[p[k]]) ll[i]++;
  }
  for (int i = 1; i <= n; i++) {
    int cnt = 0;
    for (int k = head[i]; k; k = nex[k])
      if (!lian[p[k]] && d[p[k]] - min(2, ll[p[k]]) > 1) cnt++;
    if (cnt > 2) return puts("No") & 0;
  }
  puts("Yes");
  return 0;
}
