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> v[100000];
map<int, bool> mp[100000][4];
bool dfs(int x, int y, int z) {
  int i;
  if (mp[x][z].count(y)) return mp[x][z][y];
  if (v[x].size() == 1) return mp[x][z][y] = true;
  if (z <= 1) {
    int c = 0;
    for (i = 0; i < v[x].size(); i++) {
      if (v[x][i] == y) continue;
      if (dfs(v[x][i], x, 2)) continue;
      if (!dfs(v[x][i], x, 1)) return mp[x][z][y] = false;
      c++;
    }
    if (c <= 2 - z) {
      return mp[x][z][y] = true;
    } else {
      return mp[x][z][y] = false;
    }
  } else if (z == 2) {
    if (v[x].size() > 3) return mp[x][z][y] = false;
    for (i = 0; i < v[x].size(); i++) {
      if (v[x][i] == y) continue;
      if (!dfs(v[x][i], x, 3)) return mp[x][z][y] = false;
    }
    return mp[x][z][y] = true;
  } else if (z == 3) {
    if (v[x].size() > 2) return mp[x][z][y] = false;
    for (i = 0; i < v[x].size(); i++) {
      if (v[x][i] == y) continue;
      return mp[x][z][y] = dfs(v[x][i], x, 3);
    }
  }
}
int main() {
  int n, c = 0, i;
  scanf("%d", &n);
  for (i = 0; i < n - 1; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    x--;
    y--;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  if (n <= 2) {
    puts("Yes");
    return 0;
  }
  for (i = 0; i < n; i++) {
    if (v[i].size() > v[c].size()) c = i;
  }
  if (v[c].size() > 3) {
    if (dfs(c, -1, 0)) {
      puts("Yes");
    } else {
      puts("No");
    }
    return 0;
  }
  for (i = 0; i < n; i++) {
    if (v[i].size() == 1) continue;
    if (dfs(i, -1, 0)) {
      puts("Yes");
      return 0;
    }
  }
  puts("No");
  return 0;
}
