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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const long double PI = 3.14159265358979323846;
const long double gammama = 0.57721566490153286060;
const long double eps = 1e-5;
const int INF = 1000 * 1000 * 1000 + 1;
const long long LINF = (long long)1000 * 1000 * 1000 * 1000 * 1000 * 1000;
const long long mod = 1000 * 1000 * 1000 + 7;
const int N = 500;
int n;
vector<vector<int> > g;
vector<int> d;
vector<vector<int> > dd;
void dfs(int pos, int pr = -1) {
  d[pos] = 0;
  for (int i = 0; i < 6; ++i) dd[pos][i] = 0;
  for (int i = 0; i < g[pos].size(); ++i) {
    int to = g[pos][i];
    if (to == pr) continue;
    dfs(to, pos);
    dd[pos][d[to]]++;
  }
  if (dd[pos][5] > 0) {
    d[pos] = 5;
    return;
  }
  if (dd[pos][4] > 0) {
    if (dd[pos][4] > 1) {
      d[pos] = 5;
      return;
    }
    for (int i = 1; i < 4; ++i) {
      if (dd[pos][i] > 0) {
        d[pos] = 5;
        return;
      }
    }
    if (dd[pos][0] == 0) {
      d[pos] = 4;
      return;
    }
    if (dd[pos][0] > 1) {
      d[pos] = 5;
      return;
    }
    if (pr != -1) {
      d[pos] = 5;
      return;
    }
    d[pos] = 4;
    return;
  }
  if (dd[pos][3] > 0) {
    int x = 0;
    for (int i = 1; i < 4; ++i) x += dd[pos][i];
    if (x > 1) {
      d[pos] = 5;
      return;
    }
    if (dd[pos][0] > 2) {
      d[pos] = 5;
      return;
    }
    if (dd[pos][0] <= 1) {
      d[pos] = 4;
      return;
    }
    if (pr == -1) {
      d[pos] = 4;
    } else
      d[pos] = 5;
    return;
  }
  if (dd[pos][2] > 0) {
    if (dd[pos][2] > 2) {
      d[pos] = 5;
      return;
    }
    if (dd[pos][2] == 2) {
      d[pos] = 3;
      return;
    }
    d[pos] = 2;
    return;
  }
  if (dd[pos][1] > 0) {
    d[pos] = 2;
    return;
  }
  if (dd[pos][0] <= 1) {
    d[pos] = 0;
    return;
  }
  if (dd[pos][0] == 2) {
    d[pos] = 1;
    return;
  }
  d[pos] = 2;
  return;
}
void solve() {
  cin >> n;
  g.resize(n);
  dd.assign(n, vector<int>(6, 0));
  for (int i = 0; i + 1 < n; ++i) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  d.resize(n);
  dfs(0);
  if (d[0] == 5)
    cout << "No\n";
  else
    cout << "Yes\n";
}
int main() {
  solve();
  return 0;
}
