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 MAX = 100020;
int n, edgeCount[MAX];
vector<int> to[MAX];
void input() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    int f, s;
    scanf("%d%d", &f, &s);
    edgeCount[f]++;
    edgeCount[s]++;
    to[f].push_back(s);
    to[s].push_back(f);
  }
}
priority_queue<pair<int, int> > pq;
int type[MAX];
void init() {
  for (int node = 1; node <= n; node++) {
    if (edgeCount[node] == 1) {
      type[node] = 1;
      pq.push(pair<int, int>(-type[node], node));
      edgeCount[node]--;
    }
  }
}
bool popped[MAX];
void findType(int node) {
  int all = 0, line = 0, tshape = 0, other = 0;
  for (int edge = 0; edge < to[node].size(); edge++) {
    int target = to[node][edge];
    if (popped[target]) {
      all++;
      if (type[target] == 1)
        line++;
      else if (type[target] == 2)
        tshape++;
      else if (type[target] == 3)
        other++;
    }
  }
  if (all == line) {
    if (line == 1)
      type[node] = 1;
    else if (line == 2)
      type[node] = 2;
    else
      type[node] = 3;
  } else if (other <= 1) {
    type[node] = 3;
  } else if (other == 2) {
    type[node] = 4;
  } else {
    type[node] = 5;
  }
}
bool ans = true;
void solve() {
  int countNode = 0;
  while (!pq.empty()) {
    int now = pq.top().second;
    pq.pop();
    countNode++;
    popped[now] = 1;
    int next = 0;
    for (int edge = 0; edge < to[now].size(); edge++) {
      int target = to[now][edge];
      if (edgeCount[target] != 0) {
        next = target;
        edgeCount[target]--;
      }
    }
    if (edgeCount[next] == 1) {
      findType(next);
      if ((type[next] == 4 && countNode != n - 1) || type[next] == 5) {
        ans = false;
        return;
      }
      pq.push(pair<int, int>(-type[next], next));
      edgeCount[next]--;
    }
  }
}
int main() {
  input();
  init();
  if (n > 2) {
    solve();
  }
  puts(ans ? "Yes" : "No");
  return 0;
}
