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;
int N, deg[100010], cnt[100010];
;
vector<int> g[100010];
bool v[100010];
bool ans = 1;
void dfs1(int nod, int p) {
  int i;
  for (i = 0; i < g[nod].size(); i++)
    if (g[nod][i] != p) dfs1(g[nod][i], nod);
  cnt[nod] = 0;
  for (i = 0; i < g[nod].size(); i++) {
    if (g[nod][i] != p) {
      cnt[nod] += cnt[g[nod][i]];
      if (deg[g[nod][i]] > 2) cnt[nod]++;
    }
  }
}
void dfs2(int nod, int p) {
  int aux1 = cnt[nod], aux2 = cnt[p];
  if (!ans) return;
  cnt[p] -= cnt[nod];
  if (deg[nod] > 2) cnt[p]--;
  cnt[nod] += cnt[p];
  if (deg[p] > 2) cnt[nod]++;
  int i, nr = 0;
  for (i = 0; i < g[nod].size(); i++) {
    if (deg[g[nod][i]] > 3 || cnt[g[nod][i]]) nr++;
  }
  if (nr > 2) {
    ans = 0;
    return;
  }
  for (i = 0; i < g[nod].size(); i++)
    if (g[nod][i] != p) {
      dfs2(g[nod][i], nod);
      if (!ans) return;
    }
  cnt[nod] = aux1;
  cnt[p] = aux2;
}
int main() {
  cin >> N;
  int i, x, y;
  for (i = 1; i < N; i++) {
    cin >> x >> y;
    deg[x]++, deg[y]++;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs1(1, 0);
  int nod = 1, nr = 0;
  for (i = 0; i < g[nod].size(); i++)
    if (deg[g[nod][i]] > 3 || cnt[g[nod][i]]) {
      nr++;
    }
  if (nr > 2) {
    ans = 0;
  }
  for (i = 0; i < g[1].size(); i++) dfs2(g[1][i], 1);
  if (ans)
    cout << "Yes";
  else
    cout << "No";
  cout << "\n";
  return 0;
}
