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;
int n, head[100005], tot, de[100005], le[100005], lf[100005];
struct node {
  int to, next;
} edge[100005 * 2];
void add_edge(int u, int v) {
  de[u]++;
  edge[tot].to = v;
  edge[tot].next = head[u];
  head[u] = tot++;
}
void dfs(int cnt, int pre) {
  lf[cnt] = 1;
  for (int i = head[cnt]; i != -1; i = edge[i].next) {
    int j = edge[i].to;
    if (j == pre) continue;
    if (de[j] <= 2) dfs(j, cnt);
  }
}
int main() {
  while (cin >> n) {
    tot = 0;
    memset(head, -1, sizeof(head));
    memset(de, 0, sizeof(de));
    memset(lf, 0, sizeof(lf));
    memset(le, 0, sizeof(le));
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      add_edge(u, v);
      add_edge(v, u);
    }
    for (int i = 1; i <= n; i++) {
      if (de[i] == 1) dfs(i, 0);
    }
    for (int i = 1; i <= n; i++) {
      for (int j = head[i]; j != -1; j = edge[j].next) {
        int k = edge[j].to;
        if (lf[k]) le[i]++;
      }
    }
    int flag = 0;
    for (int i = 1; i <= n && !flag; i++) {
      if (lf[i]) continue;
      int cn = 0;
      for (int j = head[i]; j != -1; j = edge[j].next) {
        int k = edge[j].to;
        if (lf[k] == 0 && de[k] - min(2, le[k]) >= 2) cn++;
        if (cn > 2) {
          flag = 1;
          break;
        }
      }
    }
    if (flag)
      cout << "NO\n";
    else
      cout << "YES\n";
  }
  return 0;
}
