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;
struct node {
  int to, next;
  node() { to = next = 0; }
} e[1000020];
int cnt, head[1000020], num[1000020];
void add(int u, int v) {
  ++cnt;
  e[cnt].to = v;
  e[cnt].next = head[u];
  head[u] = cnt;
}
void solve(int u, int fa) {
  int cx[10];
  memset(cx, 0, sizeof(cx));
  int sum = 0;
  memset(cx, 0, sizeof cx);
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].to;
    if (v == fa) continue;
    sum++;
    solve(v, u);
    cx[num[v]]++;
  }
  if (cx[5] || cx[4] || cx[3] > 2) {
    num[u] = 5;
    return;
  }
  if (cx[3] == 2) {
    num[u] = 4;
    return;
  }
  if (cx[3] == 1 || cx[2] >= 1 || cx[1] >= 3) {
    num[u] = 3;
    return;
  }
  if (cx[1] == 2) {
    num[u] = 2;
    return;
  }
  if (cx[1] == 1) {
    num[u] = 1;
    return;
  }
  num[u] = 1;
}
int flag;
void dfs(int x) {
  solve(x, x);
  if (num[x] <= 4) flag = 1;
}
int main() {
  int n;
  cin >> n;
  srand((unsigned)time(NULL));
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    add(x, y);
    add(y, x);
  }
  dfs(1);
  for (int i = 1; i <= 15; i++) dfs(rand() % n + 1);
  if (flag == 1)
    puts("Yes");
  else
    puts("No");
}
