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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
struct edge {
  int v, next;
} e[200010];
int ecnt = 0, head[100010];
int n;
int d[100010];
int leg[100010];
void addedge(int a, int b) {
  e[++ecnt].v = b;
  e[ecnt].next = head[a];
  head[a] = ecnt;
}
void del(int x) {
  while (d[x] <= 2) {
    d[x] = 0;
    int xx = x;
    for (int i = head[x]; ~i; i = e[i].next) {
      if (d[e[i].v]) {
        x = e[i].v;
        break;
      }
    }
    if (x == xx) break;
  }
  leg[x]++;
}
bool solve() {
  for (int i = (1); i <= (n); i++) {
    if (d[i]) {
      int cnt = 0;
      for (int j = head[i]; ~j; j = e[j].next) {
        int v = e[j].v;
        if (d[v]) {
          if (leg[v] > 2)
            cnt++;
          else if (d[v] - leg[v] > 1)
            cnt++;
        }
      }
      if (cnt > 2) return false;
    }
  }
  return true;
}
int main() {
  cin >> n;
  memset((head), (-1), sizeof(head));
  memset((d), (0), sizeof(d));
  memset((leg), (0), sizeof(leg));
  for (int i = (1); i <= (n - 1); i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    addedge(u, v);
    addedge(v, u);
    d[u]++;
    d[v]++;
  }
  for (int i = (1); i <= (n); i++) {
    if (d[i] == 1) del(i);
  }
  puts(solve() ? "Yes" : "No");
  return 0;
}
