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;
const long long inf = 1e18;
const double pi = acos(-1.0);
const int N = 300010;
int n;
bool dp[N], dp1[N], dp2[N];
vector<int> g[N];
pair<int, int> p1[N];
void dfs(int v, int u, int p) {
  for (int i = 0; i < g[v].size(); i++)
    if (g[v][i] != p) dfs(g[v][i], u, v);
  int r = 0;
  for (int i = 0; i < g[v].size(); i++)
    if (g[v][i] != p) r++;
  if (r == 0) {
    dp1[v] = true;
    dp2[v] = true;
    dp[v] = true;
    return;
  }
  if (r > 1)
    dp1[v] = false;
  else {
    for (int i = 0; i < g[v].size(); i++)
      if (g[v][i] != p) dp1[v] = dp1[g[v][i]];
  }
  if (r > 2)
    dp2[v] = false;
  else {
    dp2[v] = true;
    for (int i = 0; i < g[v].size(); i++)
      if (g[v][i] != p) {
        if (!dp1[g[v][i]]) dp2[v] = false;
      }
  }
  int k = 0, l = 0;
  for (int i = 0; i < g[v].size(); i++)
    if (g[v][i] != p) {
      if (dp2[g[v][i]])
        k++;
      else if (dp[g[v][i]])
        l++;
      else {
        dp[v] = false;
        return;
      }
    }
  if (dp1[v] || dp2[v])
    dp[v] = true;
  else {
    if (v == u) {
      if (l <= 2 && k >= r - l)
        dp[v] = true;
      else
        dp[v] = false;
    } else {
      if (l <= 1 && k >= r - l)
        dp[v] = true;
      else
        dp[v] = false;
    }
  }
}
bool solve(int v) {
  for (int i = 1; i <= n; i++) {
    dp1[i] = false;
    dp2[i] = false;
    dp[i] = false;
  }
  dfs(v, v, -1);
  return dp[v];
}
int main() {
  srand(time(0));
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (int i = 1; i <= n; i++) p1[i] = make_pair(g[i].size(), i);
  sort(p1 + 1, p1 + n + 1);
  for (int i = 1; i <= min(n, 30); i++)
    if (solve(p1[i].second)) {
      cout << "Yes";
      return 0;
    }
  for (int i = n; i >= max(1, n - 30); i--)
    if (solve(p1[i].second)) {
      cout << "Yes";
      return 0;
    }
  for (int i = 1; i <= 30; i++) {
    int v = rand() % n + 1;
    if (solve(v)) {
      cout << "Yes";
      return 0;
    }
  }
  cout << "No";
  return 0;
}
