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 N = 1e5 + 10;
vector<int> g[N];
int n;
bool deg[N];
bool Can(int v, int par) {
  int cnt = 0;
  bool ret = true;
  for (auto u : g[v])
    if (u != par) ret &= deg[u], cnt++;
  if (!ret or cnt > 2) return false;
  return true;
}
bool dfs1(int v, int par) {
  bool ret = true;
  int cnt = 0;
  for (auto u : g[v])
    if (u != par) ret &= dfs1(u, v), cnt++;
  if (cnt <= 1 && ret)
    deg[v] = true;
  else
    deg[v] = false;
  return deg[v];
}
void dfs2(int v, int par, int could, bool op) {
  int cnt = 0, ccnt = 0, cnt2 = 0;
  for (auto u : g[v]) {
    if (u == par) continue;
    if (Can(u, v)) ccnt++;
    if (deg[u]) cnt2++;
    cnt++;
  }
  if (cnt - ccnt > could) {
    cout << "No" << endl;
    exit(0);
  }
  for (auto u : g[v]) {
    if (u == par or Can(u, v)) continue;
    if (could == 1 or cnt - ccnt > 1 or ccnt != cnt2 or ccnt > 2) {
      dfs2(u, v, 1, true);
      continue;
    }
    if (par == -1) {
      if (cnt2 <= 1)
        dfs2(u, v, could, false);
      else
        dfs2(u, v, could, true);
      continue;
    }
    if (op or cnt2 > 1)
      dfs2(u, v, 1, true);
    else
      dfs2(u, v, 2, false);
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs1(0, -1);
  dfs2(0, -1, 2, false);
  cout << "Yes" << endl;
}
