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 NMax = 110000;
struct edge {
  int num, t;
  edge *next;
} * G[NMax], pool[NMax * 6];
int N, del[NMax], D[NMax], L;
void Build(int x, int y, int z = 0) {
  edge *p = &pool[L++], *q = &pool[L++];
  p->num = y;
  p->t = z;
  p->next = G[x];
  G[x] = p;
  q->num = x;
  q->t = z;
  q->next = G[y];
  G[y] = q;
}
int size[NMax], cnt1[NMax];
void DFS1(int a, int fa) {
  size[a] = 1;
  for (edge *p = G[a]; p; p = p->next)
    if (!del[p->num] && p->num != fa) {
      DFS1(p->num, a);
      size[a] += size[p->num];
      if (size[p->num] > 3 || (size[p->num] == 3 && p->t)) cnt1[a]++;
      if (N - size[p->num] > 3 || (N - size[p->num] == 3 && p->t))
        cnt1[p->num]++;
    }
}
int DFS2(int a, int fa, int t) {
  if (t == 0) {
    int cnt = 0;
    for (edge *p = G[a]; p; p = p->next)
      if (!del[p->num] && p->num != fa) {
        if (size[p->num] > 3 || (size[p->num] == 3 && p->t)) cnt++;
      }
    if (cnt > 2) return 0;
    for (edge *p = G[a]; p; p = p->next)
      if (!del[p->num] && p->num != fa) {
        if (size[p->num] > 3 || (size[p->num] == 3 && p->t)) {
          int x = DFS2(p->num, a, 1);
          if (!x) return 0;
        }
      }
    return 1;
  } else {
    int cnt = 0;
    for (edge *p = G[a]; p; p = p->next)
      if (!del[p->num] && p->num != fa) {
        if (size[p->num] > 3 || (size[p->num] == 3 && p->t)) cnt++;
      }
    if (cnt > 1) return 0;
    for (edge *p = G[a]; p; p = p->next)
      if (!del[p->num] && p->num != fa) {
        if (size[p->num] > 3 || (size[p->num] == 3 && p->t)) {
          int x = DFS2(p->num, a, 1);
          if (!x) return 0;
        }
      }
    return 1;
  }
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i < N; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    D[x]++;
    D[y]++;
    Build(x, y);
  }
  for (int i = 1; i <= N; i++)
    if (D[i] == 2 && !del[i]) {
      int x = -1, y = -1;
      for (edge *p = G[i]; p; p = p->next)
        if (!del[p->num]) {
          if (x == -1)
            x = p->num;
          else
            y = p->num;
        }
      Build(x, y, 1);
      del[i] = 1;
    }
  int R = 1, maxx1 = 0;
  for (int i = 1; i <= N; i++)
    if (!del[i]) R = i;
  DFS1(R, -1);
  for (int i = 1; i <= N; i++)
    if (!del[i]) {
      if (cnt1[i] > maxx1) {
        maxx1 = cnt1[i];
        R = i;
      }
    }
  DFS1(R, -1);
  if (DFS2(R, -1, 0))
    puts("Yes");
  else
    puts("No");
  getchar();
  getchar();
  return 0;
}
