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;
int n, deg[N], flag, sum[N][6], num[N];
vector<int> g[N];
int calc(int x) {
  if (sum[x][5] || sum[x][3] > 2 || sum[x][4]) return 5;
  if (sum[x][3] > 1) return 4;
  if (sum[x][3] || sum[x][2] || sum[x][1] > 2) return 3;
  if (sum[x][1] > 1) return 2;
  return 1;
}
void dfs(int x, int fa) {
  for (int i = 0; i < g[x].size(); i++) {
    int y = g[x][i];
    if (y == fa) continue;
    dfs(y, x);
    sum[x][num[y]]++;
  }
  num[x] = calc(x);
}
void dfs2(int x, int fa, int v) {
  sum[x][v]++;
  if (calc(x) < 5) {
    flag = 1;
    return;
  }
  for (int i = 0; i < g[x].size(); i++) {
    int y = g[x][i];
    if (y == fa) continue;
    sum[x][num[y]]--;
    dfs2(y, x, calc(x));
    if (flag) return;
    sum[x][num[y]]++;
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n - 1; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    g[x].push_back(y), g[y].push_back(x);
    ++deg[x], ++deg[y];
  }
  dfs(1, 0);
  dfs2(1, 0, 0);
  puts(flag ? "Yes" : "No");
  return 0;
}
