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 = (int)1e5 + 100;
vector<int> g[maxn], G[maxn];
int type[maxn];
int cntIn[maxn];
void no() {
  printf("No");
  exit(0);
}
void yes() {
  printf("Yes");
  exit(0);
}
void dfsG(int v, int par = -1) {
  if ((int)G[v].size() > 2) no();
  type[v] = 3;
  for (int i = 0; i < (int)G[v].size(); i++) {
    int nv = G[v][i];
    if (nv == par) continue;
    dfsG(nv, v);
    break;
  }
}
void solve() {
  int n;
  scanf("%d", &n);
  if (n <= 2) yes();
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  bool fnd = false;
  for (int i = 0; i < n; i++)
    if ((int)g[i].size() >= 3) fnd = true;
  if (!fnd) yes();
  for (int i = 0; i < n; i++)
    if ((int)g[i].size() == 1 && type[i] == 0) {
      type[i] = 1;
      int pv = i;
      int v = g[i][0];
      while ((int)g[v].size() == 2) {
        type[v] = 1;
        int nv = g[v][0] + g[v][1] - pv;
        pv = v;
        v = nv;
      }
      cntIn[v]++;
    }
  for (int i = 0; i < n; i++)
    if ((int)g[i].size() == 3 && cntIn[i] == 2) {
      type[i] = 2;
    }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < (int)g[i].size(); j++) {
      int ni = g[i][j];
      if (type[i] == 0 && type[ni] == 0) G[i].push_back(ni);
    }
  int st = -1;
  for (int i = 0; i < n; i++)
    if (type[i] == 0 && (int)G[i].size() <= 2) st = i;
  if (st != -1) {
    type[st] = 3;
    for (int i = 0; i < (int)G[st].size(); i++) dfsG(G[st][i], st);
  }
  for (int i = 0; i < n; i++)
    if (type[i] == 0) no();
  yes();
}
int main(int, char **) {
  solve();
  return 0;
}
