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 branch[100010];
bool ok;
vector<int> g[100010];
void dfs(int u, int fa) {
  for (int v : g[u]) {
    if (v == fa) continue;
    dfs(v, u);
    branch[u] += branch[v];
  }
  if ((int)g[u].size() > 2) branch[u]++;
}
void dfs2(int u, int fa, int branchCnt) {
  int cnt = 0;
  if (branchCnt) cnt++;
  for (int v : g[u]) {
    if (v == fa) continue;
    dfs2(v, u, branchCnt + branch[u] - branch[v]);
    if (branch[v]) cnt++;
  }
  if (branchCnt == 1 && fa != -1 && (int)g[fa].size() == 3) cnt--;
  for (int v : g[u]) {
    if (v == fa) continue;
    if (branch[v] == 1 && (int)g[v].size() == 3) cnt--;
  }
  if (cnt > 2) ok = false;
}
int main() {
  int n, a, b;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  ok = true;
  dfs(1, -1);
  dfs2(1, -1, 0);
  if (ok)
    puts("Yes");
  else
    puts("No");
  return 0;
}
