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;
vector<int> edges[100008];
bool vis[100008], leaf[100008];
void dfs(int u, int fa) {
  if ((int)edges[u].size() <= 2) {
    vis[u] = true;
    for (int i = 0; i < (int)edges[u].size(); i++)
      if (edges[u][i] != fa) dfs(edges[u][i], u);
  }
  return;
}
int main() {
  int n, i, j;
  bool ans = true;
  scanf("%d", &n);
  for (i = 0; i < n; i++) edges[i].clear();
  for (i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    --u;
    --v;
    edges[u].push_back(v);
    edges[v].push_back(u);
  }
  memset(vis, 0, sizeof(vis));
  for (i = 0; i < n; i++)
    if ((int)edges[i].size() == 1) dfs(i, -1);
  memset(leaf, 0, sizeof(leaf));
  for (i = 0; i < n; i++)
    if (!vis[i]) {
      int cnt = 0;
      for (j = 0; j < (int)edges[i].size(); j++) {
        int u = edges[i][j];
        if (!vis[u]) ++cnt;
      }
      if (cnt == 1 && (int)edges[i].size() <= 3) leaf[i] = true;
    }
  for (i = 0; i < n && ans; i++)
    if (!vis[i]) {
      int cnt1 = 0, cnt2 = 0;
      for (j = 0; j < (int)edges[i].size(); j++) {
        int u = edges[i][j];
        if (!vis[u]) {
          ++cnt1;
          if (leaf[u]) ++cnt2;
        }
      }
      if (cnt1 - cnt2 > 2) ans = false;
    }
  if (ans)
    puts("Yes");
  else
    puts("No");
  return 0;
}
