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 y, t;
} e[200002];
bool vis[100002];
char ch;
int n, x, y, h[100002], ep, sz[100002], d[100002], w;
inline void read(int &x) {
  x = 0;
  do {
    ch = getchar();
  } while (ch < '0' || '9' < ch);
  do {
    x = x * 10 + ch - '0';
    ch = getchar();
  } while ('0' <= ch && ch <= '9');
}
inline void AddEdge(const int &x, const int &y) {
  e[++ep].y = y;
  e[ep].t = h[x];
  h[x] = ep;
}
void dfs(const int &x) {
  vis[x] = true;
  for (int i = h[x]; i; i = e[i].t)
    if (!vis[e[i].y] && sz[e[i].y] <= 2) dfs(e[i].y);
}
int main(void) {
  read(n);
  for (int i = 1; i < n; i++)
    read(x), read(y), AddEdge(x, y), AddEdge(y, x), sz[x]++, sz[y]++;
  for (int i = 1; i <= n; i++)
    if (sz[i] == 1) dfs(i);
  for (int i = 1; i <= n; i++)
    if (!vis[i]) {
      for (int j = h[i]; j; j = e[j].t)
        if (vis[e[j].y]) {
          d[i] = min(d[i] + 1, 2);
        }
    }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) {
      w = 0;
      for (int j = h[i]; j; j = e[j].t)
        if (!vis[e[j].y] && sz[e[j].y] - d[e[j].y] > 1) w++;
      if (w > 2) {
        puts("No");
        return 0;
      }
    }
  puts("Yes");
  return 0;
}
