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 = 200000 + 10;
int n, deg[maxn];
vector<int> edge[maxn];
int on[maxn], q[maxn];
void nosol() {
  printf("No\n");
  exit(0);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) deg[i] = 0, edge[i].clear(), on[i] = 1;
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    edge[a].push_back(b);
    edge[b].push_back(a);
    deg[a]++, deg[b]++;
  }
  int hd = 0, tl = 0;
  for (int i = 1; i <= n; i++)
    if (deg[i] == 1) q[++tl] = i;
  while (hd != tl) {
    int x = q[++hd];
    on[x] = 0;
    for (int i = 0; i < (int)edge[x].size(); i++) {
      int y = edge[x][i];
      if (!on[y]) continue;
      deg[y]--;
      if (deg[y] == 1 && (int)edge[y].size() <= 2) q[++tl] = y;
    }
  }
  for (int i = 1; i <= n; i++)
    if (on[i]) {
      if (deg[i] == 1 && (int)edge[i].size() == 3) on[i] = 0;
    }
  for (int i = 1; i <= n; i++)
    if (on[i]) {
      int degree = 0;
      for (int j = 0; j < (int)edge[i].size(); j++)
        if (on[edge[i][j]]) degree++;
      if (degree > 2) nosol();
    }
  printf("Yes\n");
  return 0;
}
