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;
const int inf = ~0U >> 1;
const long long INF = ~0ULL >> 1;
template <class T>
inline void read(T &n) {
  char c;
  for (c = getchar(); !(c >= '0' && c <= '9'); c = getchar())
    ;
  n = c - '0';
  for (c = getchar(); c >= '0' && c <= '9'; c = getchar()) n = n * 10 + c - '0';
}
const int maxn = 200000;
vector<int> E[maxn];
int d[maxn], onPath[maxn], x, y, N;
queue<int> Q;
int main() {
  scanf("%d", &N);
  for (int i = (1); i <= (N - 1); ++i)
    scanf("%d%d", &x, &y), E[x].push_back(y), E[y].push_back(x), d[x]++, d[y]++;
  for (int i = (1); i <= (N); ++i) onPath[i] = 1;
  for (int i = (1); i <= (N); ++i)
    if (d[i] == 1) Q.push(i);
  while (Q.size()) {
    int u = Q.front();
    Q.pop();
    onPath[u] = 0;
    for (auto v : E[u]) {
      --d[v];
      if (d[v] == 1 && E[v].size() <= 2) Q.push(v);
    }
  }
  for (int i = (1); i <= (N); ++i)
    if (onPath[i] && E[i].size() == 3 && d[i] == 1) onPath[i] = 0;
  for (int i = (1); i <= (N); ++i) {
    int cnt = 0;
    for (auto j : E[i])
      if (onPath[j]) ++cnt;
    if (cnt > 2) {
      puts("No");
      return 0;
    }
  }
  puts("Yes");
  return 0;
}
