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;
long long int mod = 1e9 + 7;
int a[1000100], s[100100];
vector<int> v[100100];
bool fl;
int dfs(int t) {
  int i, j = 0, ans = 0, d = 0, l = 0, e1 = 0, e2 = 0;
  a[t] = 1;
  for (i = 0; i < v[t].size(); i++)
    if (a[v[t][i]] == 0) {
      j = dfs(v[t][i]);
      if (j == 0) l++;
      if (j == 1) d++;
      if (j == 2) e1++;
      if (j == 3) e2++;
      if (j > 3) {
        fl = false;
        return 4;
      }
    }
  if (e2 * 2 + e1 > 2) {
    fl = false;
    return 4;
  }
  if (e2 == 1) {
    if (v[t].size() > 3 || d > 0) {
      fl = false;
      return 4;
    }
    return 3;
  }
  if (e1 >= 1) {
    return e1 + 1;
  }
  if (e1 == 0) {
    if (d > 0 || v[t].size() > 3) return 2;
  }
  if (v[t].size() >= 3) return 1;
  return 0;
}
void solve(int n) {
  int i, j, l, q = 0, w = 0, z, r, k = 0, t;
  for (i = 0; i < n; i++) {
    v[i].clear();
    a[i] = 0;
  }
  for (i = 0; i < n - 1; i++) {
    scanf("%d%d", &q, &w);
    q--;
    w--;
    v[q].push_back(w);
    v[w].push_back(q);
  }
  j = -1;
  for (i = 0; i < n; i++)
    if (v[i].size() > 2) {
      j = i;
    }
  fl = true;
  if (j >= 0) dfs(j);
  if (fl)
    printf("Yes\n");
  else
    printf("No\n");
}
int main() {
#pragma comment(linker, "/STACK:1073741824")
  int n, m, k, i;
  while (scanf("%d", &n) != EOF) solve(n);
  return 0;
}
