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>
const bool DEBUG = false;
std::ifstream fin("in");
std::ofstream fout("out");
using std::cerr;
using std::cin;
using std::cout;
using std::endl;
using std::istream;
using std::ostream;
const int INF = 1 << 30;
const int MAXV = 100000, D = 500;
struct edge {
  int y, next;
  static int tot;
} e[MAXV * 2 + D];
int edge::tot;
int d[MAXV + D], head[MAXV + D];
bool leaf[MAXV + D];
bool son[MAXV + D];
inline void connect(int x, int y) {
  d[x]++, d[y]++;
  int& k = edge::tot;
  e[++k].y = y, e[k].next = head[x], head[x] = k;
  e[++k].y = x, e[k].next = head[y], head[y] = k;
}
int n;
void init(istream& in) {
  in >> n;
  for (int i = 1; i <= n - 1; i++) {
    int x, y;
    in >> x >> y;
    connect(x, y);
  }
}
void dfs(int x) {
  leaf[x] = true;
  for (int i = head[x]; i; i = e[i].next) {
    if (leaf[e[i].y] == true) continue;
    if (d[e[i].y] == 2) dfs(e[i].y);
  }
}
void work() {
  for (int i = 1; i <= n; i++)
    if (d[i] == 1) dfs(i);
  for (int x = 1; x <= n; x++) {
    if (d[x] == 3) {
      int p = 3;
      for (int i = head[x]; i; i = e[i].next)
        if (leaf[e[i].y]) p--;
      if (p == 1) son[x] = true;
    }
  }
  for (int x = 1; x <= n; x++) {
    if (son[x] || leaf[x])
      for (int i = head[x]; i; i = e[i].next) d[e[i].y]--;
  }
  for (int i = 1; i <= n; i++)
    if (son[i] == false && leaf[i] == false && d[i] > 2) {
      cout << "No" << endl;
      return;
    }
  cout << "Yes" << endl;
}
int main() {
  std::ios::sync_with_stdio(false);
  init(DEBUG ? fin : cin);
  work();
  if (DEBUG) cerr << "Time has passed " << clock() * 1.0 / 1000 << "s.\n";
  return 0;
}
