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 = (int)1e9 + 10;
const long long LLINF = (long long)1e18 + 10;
const double EPS = 1e-12;
struct edge {
  int to, nxt;
  edge() {
    nxt = 0;
    to = 0;
  }
  edge(int to, int nxt) : to(to), nxt(nxt) {}
};
const int N = 100005;
edge e[N + N];
int h[N], deg[N], last = 1, n;
int I[N + N], Y[N + N];
inline void add(int from, int to) {
  last++;
  deg[from]++;
  e[last] = edge(to, h[from]);
  h[from] = last;
}
inline int linear(int eid) {
  if (I[eid] != -1) return I[eid];
  int u = e[eid].to;
  int childs = deg[u] - 1;
  if (childs == 0) return I[eid] = true;
  if (childs == 1) {
    int nxt = -1;
    for (int i = h[u]; i; i = e[i].nxt)
      if (i != (eid ^ 1)) nxt = i;
    assert(nxt != -1);
    return I[eid] = linear(nxt);
  }
  return I[eid] = false;
}
inline int midchild(int eid) {
  if (Y[eid] != -1) return Y[eid];
  int u = e[eid].to;
  int childs = deg[u] - 1;
  if (childs <= 2) {
    for (int i = h[u]; i; i = e[i].nxt) {
      if ((i == (eid ^ 1))) continue;
      if (!linear(i)) return Y[eid] = false;
    }
    return Y[eid] = true;
  }
  return Y[eid] = false;
}
inline int maincheck(int u) {
  int badchilds = 0;
  for (int i = h[u]; i; i = e[i].nxt) {
    badchilds += int(!midchild(i));
  }
  return badchilds <= 2;
}
int main() {
  memset(h, 0, sizeof h);
  memset(deg, 0, sizeof deg);
  memset(I, -1, sizeof I);
  memset(Y, -1, sizeof Y);
  scanf("%d", &n);
  for (int i = 0; i <= (int)(n - 2); ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  for (int i = 1; i <= n; ++i)
    if (!maincheck(i)) {
      puts("No");
      return 0;
    }
  puts("Yes");
}
