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;
inline long long RR() {
  long long v = 0, k = 1;
  char c = 0;
  while ('0' > c || c > '9') {
    if (c == '-') k = -1;
    c = getchar();
  }
  while ('0' <= c && c <= '9') v = v * 10 + c - 48, c = getchar();
  return v * k;
}
inline void W(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) W(x / 10);
  putchar(x % 10 + 48);
}
inline void WW(long long x) {
  W(x);
  puts("");
}
inline void read(int &x, int &y) {
  x = RR();
  y = RR();
}
const int N = 1e5 + 5;
vector<int> h[N];
int n, m, x, y, S;
int A[N], d[N];
void dfs1(int x, int y) {
  if (d[x] == 1) return;
  if (h[x].size() > 2) return;
  d[x] = 1;
  for (int k = 0; k < h[x].size(); k++)
    if (h[x][k] != y) dfs1(h[x][k], x);
}
void die() {
  puts("No");
  exit(0);
}
void dfs2(int x) {
  d[x] = 3;
  int sum = 0;
  for (int k = 0; k < h[x].size(); k++)
    if (d[h[x][k]] == 0) sum++;
  if (sum > 1) die();
  for (int k = 0; k < h[x].size(); k++)
    if (d[h[x][k]] == 0) dfs2(h[x][k]);
}
void check(int x) {
  if (h[x].size() != 3) return;
  int ans = 0;
  for (int k = 0; k < h[x].size(); k++)
    if (d[h[x][k]] == 1) ans++;
  if (ans == 2) d[x] = 2;
}
int main() {
  n = RR();
  for (int i = 1; i < n; i++) {
    read(x, y);
    h[x].push_back(y);
    h[y].push_back(x);
    A[x]++;
    A[y]++;
  }
  for (int i = 1; i <= n; i++)
    if (A[i] == 1) dfs1(i, -1);
  for (int i = 1; i <= n; i++) check(i);
  for (int i = 1; i <= n; i++)
    if (d[i] == 0) {
      int sum = 0;
      d[i] = 3;
      for (int k = 0; k < h[i].size(); k++)
        if (d[h[i][k]] == 0) sum++;
      if (sum > 2) die();
      for (int k = 0; k < h[i].size(); k++)
        if (d[h[i][k]] == 0) dfs2(h[i][k]);
    }
  puts("Yes");
}
