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;
void write(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
int read() {
  int d = 0, w = 1;
  char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') w = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) d = (d << 1) + (d << 3) + c - 48;
  return d * w;
}
void wln(int x) {
  write(x);
  putchar('\n');
}
void wrs(int x) {
  write(x);
  putchar(' ');
}
vector<int> a[100010];
int num[100010][5], sum[100010], f, n;
int calc(int x) {
  if (num[x][5] || num[x][4] || num[x][3] > 2) return 5;
  if (num[x][3] > 1) return 4;
  if (num[x][3] || num[x][2] || num[x][1] > 2) return 3;
  if (num[x][1] > 1) return 2;
  return 1;
}
void dfs1(int x, int fa) {
  for (auto v : a[x]) {
    if (v == fa) continue;
    dfs1(v, x);
    num[x][sum[v]]++;
  }
  sum[x] = calc(x);
}
void dfs2(int x, int now, int fa) {
  num[x][now]++;
  if (calc(x) < 5) {
    f = 1;
    return;
  }
  for (auto v : a[x]) {
    if (v == fa) continue;
    num[x][sum[v]]--;
    dfs2(v, calc(x), x);
    num[x][sum[v]]++;
  }
  num[x][now]--;
}
signed main() {
  n = read();
  for (int i = 1; i < n; i++) {
    int x = read(), y = read();
    a[x].push_back(y);
    a[y].push_back(x);
  }
  dfs1(1, 0);
  dfs2(1, 0, 0);
  puts(f ? "Yes" : "No");
  return 0;
}
