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;
int n, tot, e[200001], head[200001], nxt[200001], g[200001], ru[200001],
    b[200001];
void build(int t, int k) {
  tot++;
  e[tot] = k;
  nxt[tot] = head[t];
  head[t] = tot;
}
void dfs(int x) {
  int i;
  if (ru[x] <= 2) {
    g[x] = 1;
    for (i = head[x]; i; i = nxt[i])
      if (!g[e[i]]) dfs(e[i]);
  }
}
int pd() {
  int i, j, flag;
  for (i = 1; i <= n; i++)
    if (ru[i] == 1) dfs(i);
  for (i = 1; i <= n; i++)
    if (!g[i]) {
      for (j = head[i]; j; j = nxt[j])
        if (!g[e[j]]) b[i]++;
    }
  for (i = 1; i <= n; i++)
    if (ru[i] == 3 && b[i] == 1) {
      g[i] = 1;
    }
  for (i = 1; i <= n; i++)
    if (!g[i]) {
      flag = 0;
      for (j = head[i]; j; j = nxt[j])
        if (!g[e[j]]) flag++;
      if (flag > 2) return 0;
    }
  return 1;
}
int main() {
  int i, t, k;
  scanf("%d", &n);
  for (i = 1; i < n; i++) {
    scanf("%d%d", &t, &k);
    ru[t]++;
    ru[k]++;
    build(t, k);
    build(k, t);
  }
  if (pd())
    puts("Yes");
  else
    puts("No");
}
