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 warning(disable : 4996)
int cnt, now[100001], f[100001], g[100001];
bool d[100001];
struct {
  int pre, p;
} e[200001];
inline int min(const int& a, const int& b) { return a < b ? a : b; }
inline void add(int a, int b) {
  e[++cnt].pre = now[a];
  now[a] = cnt;
  e[cnt].p = b;
}
void dfs(int x, int fa = 0) {
  d[x] = 1;
  for (int i = now[x]; i; i = e[i].pre)
    if (fa != e[i].p && f[e[i].p] < 3) dfs(e[i].p, x);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    f[a]++;
    f[b]++;
    add(a, b);
    add(b, a);
  }
  for (int i = 1; i <= n; i++)
    if (f[i] == 1 && !d[i]) dfs(i);
  for (int i = 1; i <= n; i++)
    if (!d[i])
      for (int j = now[i]; j; j = e[j].pre)
        if (d[e[j].p]) g[i]++;
  for (int i = 1; i <= n; i++)
    if (!d[i]) g[i] = min(g[i], 2);
  for (int i = 1; i <= n; i++)
    if (!d[i]) {
      int cnt = 0;
      for (int j = now[i]; j; j = e[j].pre)
        if (!d[e[j].p] && f[e[j].p] - g[e[j].p] > 1) cnt++;
      if (cnt > 2) {
        puts("No");
        return 0;
      }
    }
  puts("Yes");
  return 0;
}
