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 MAXN = 100005;
const int MAXE = 200005;
int n;
struct Graph {
  int head[MAXN];
  int next[MAXE];
  int end[MAXE];
  int deg[MAXN];
  int ndeg[MAXN];
  bool del[MAXN];
  int tot;
  void init() { tot = 2; }
  void addEdge(int from, int to) {
    end[tot] = to;
    next[tot] = head[from];
    head[from] = tot++;
    deg[from]++;
  }
} gph;
void dfs(int cur, int par = 0) {
  if (gph.deg[cur] <= 2) {
    gph.del[cur] = true;
    for (int j = gph.head[cur]; j; j = gph.next[j]) {
      if (gph.end[j] != par) {
        dfs(gph.end[j], cur);
      }
    }
  }
}
bool judge() {
  for (int i = 1; i <= n; i++) {
    if (gph.deg[i] == 1) {
      dfs(i);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (gph.deg[i] >= 3) {
      int ndeg = 0;
      for (int j = gph.head[i]; j; j = gph.next[j]) {
        if (!gph.del[gph.end[j]]) {
          ndeg++;
        }
      }
      gph.ndeg[i] = ndeg;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (gph.deg[i] == 3 && gph.ndeg[i] == 1) {
      gph.del[i] = true;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (!gph.del[i]) {
      int fdeg = 0;
      for (int j = gph.head[i]; j; j = gph.next[j]) {
        if (!gph.del[gph.end[j]]) {
          fdeg++;
        }
      }
      if (fdeg > 2) {
        return false;
      }
    }
  }
  return true;
}
int main() {
  gph.init();
  scanf(" %d", &n);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf(" %d %d", &a, &b);
    gph.addEdge(a, b);
    gph.addEdge(b, a);
  }
  if (judge()) {
    cout << "Yes" << endl;
  } else {
    cout << "No" << endl;
  }
  return 0;
}
