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;
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
vector<int> adj[100100];
int dp[100100];
bool dfs(int u, int fa = 0) {
  int c[6] = {
      0,
  };
  for (int i = adj[u].size(); i--;) {
    int v = adj[u][i];
    if (v == fa) continue;
    if (!dfs(v, u)) return false;
    c[dp[v]]++;
  }
  if (!c[1] && !c[2] && !c[3] && !c[4] && !c[5]) {
    if (c[0] <= 1)
      dp[u] = 0;
    else if (c[0] == 2)
      dp[u] = 1;
    else
      dp[u] = 2;
  } else {
    if (c[5] || c[2] > 2 || c[3] > 1 || c[3] && (c[1] + c[2] || c[0] > 2) ||
        c[4] && (c[0] + c[1] + c[2] + c[3]) > 1)
      return false;
    if (c[3] && (c[0] + c[1] + c[2]) >= 2 ||
        c[4] && (c[0] + c[1] + c[2] + c[3]))
      dp[u] = 5;
    else if (c[3] && c[0] + c[1] + c[2] || c[4])
      dp[u] = 4;
    else if (c[2] > 1 || c[3])
      dp[u] = 3;
    else
      dp[u] = 2;
  }
  return true;
}
int main() {
  int n, u, v;
  cin >> n;
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  int t = 99, ans = 0;
  while (t--) {
    if (dfs(rand() % n + 1)) ans++;
  }
  puts(ans >= 97 ? "Yes" : "No");
  return 0;
}
