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 int MAXN = 101 * 1001;
vector<int> a[MAXN];
int is1[MAXN], is2[MAXN] = {}, is3[MAXN] = {}, is4[MAXN] = {};
int par[MAXN] = {}, q[MAXN] = {}, t[MAXN] = {};
void dfs(int u, int p = -1) {
  par[u] = p;
  int x = 1;
  for (auto v : a[u]) {
    if (v == p) continue;
    dfs(v, u);
    if (!is1[v]) x = 0;
  }
  int k = a[u].size();
  if (p != -1) k--;
  t[u] = k;
  if (x == 0) {
    if (p != -1) q[p]++;
    return;
  }
  if (k == 0) is1[u] = 1;
  if (k == 1) is1[u] = 1;
  if (k == 2) is2[u] = 1;
  if (k > 2 && p != -1) q[p]++;
}
void dfs2(int u, int p = -1) {
  if (p != -1) {
    int x = 0;
    if (par[p] != -1) {
      if (!is3[p])
        x = 100;
      else
        x++;
    }
    if (t[p] - 1 + x > 2)
      x = 100;
    else {
      for (auto v : a[p]) {
        if (v == par[p] || v == u) continue;
        if (!is1[v]) x = 100;
      }
    }
    if (x == 100)
      q[u]++;
    else {
      if (t[p] - 1 == 0 && is3[p]) is3[u] = 1;
    }
  } else
    is3[u] = 1;
  for (auto v : a[u]) {
    if (v == p) continue;
    dfs2(v, u);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, x, y;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> x >> y;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  for (int i = 1; i <= n; i++)
    if (a[i].size() == 1) x = i;
  dfs(x);
  dfs2(x);
  x = 0;
  for (int i = 1; i <= n; i++) {
    if (q[i] > 2) return cout << "No", 0;
  }
  cout << "Yes";
}
