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;
struct edge {
  int t, next;
} e[100005 * 2];
int ecnt, head[100005], du[100005], del[100005];
void addedge(int f, int t) {
  e[++ecnt] = (edge){t, head[f]};
  head[f] = ecnt;
  e[++ecnt] = (edge){f, head[t]};
  head[t] = ecnt;
  du[f]++;
  du[t]++;
}
bool DFS(int u, int fr) {
  int cnt = 0;
  bool ret = true;
  for (int i = head[u]; i; i = e[i].next) {
    if (e[i].t != fr && del[e[i].t] == 0) (ret &= DFS(e[i].t, u)), cnt++;
  }
  if (cnt >= 2) ret = false;
  return ret;
}
int n, cnt, buf[100005];
void tp(int u) {
  for (int i = head[u]; i; i = e[i].next) {
    if (del[e[i].t] == 0 && du[e[i].t] <= 2) {
      du[e[i].t]--;
      del[u] = 1;
      cnt--;
      tp(e[i].t);
      return;
    } else if (del[e[i].t] == 0 && du[e[i].t] == 3) {
      del[u] = 1;
      cnt--;
      buf[e[i].t]--;
      return;
    }
  }
}
int main() {
  scanf("%d", &n);
  cnt = n;
  for (int i = 1, f, t; i < n; i++) {
    scanf("%d%d", &f, &t);
    addedge(f, t);
  }
  for (int i = 1; i <= n; i++) {
    if (del[i] == 0 && du[i] == 1 && du[e[head[i]].t] <= 3) tp(i);
  }
  for (int i = 1; i <= n; i++) du[i] += buf[i];
  memset(buf, 0, sizeof(buf));
  for (int i = 1; i <= n; i++) {
    if (del[i] == 0 && du[i] == 1) {
      del[i] = 1, cnt--;
      for (int j = head[i]; j; j = e[j].next) {
        buf[e[j].t]--;
      }
    }
  }
  for (int i = 1; i <= n; i++) du[i] += buf[i];
  if (cnt <= 1) return puts("Yes"), 0;
  for (int i = 1; i <= n; i++) {
    if (del[i] == 0 && du[i] == 1) {
      if (DFS(i, 0))
        return puts("Yes"), 0;
      else
        return puts("No"), 0;
    }
  }
  return 0;
}
