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 long long MAXN = 1e5 + 10;
vector<long long> adj[MAXN];
bool dpd[MAXN], dpd2[MAXN], dpu[MAXN], dpu2[MAXN];
long long n, u, v, cntt, par[MAXN];
void dfs_down(long long v, long long p) {
  par[v] = p;
  long long cnt = 0;
  for (long long i = 0; i < (long long)adj[v].size(); i++) {
    if (adj[v][i] == p) continue;
    dfs_down(adj[v][i], v);
    if (dpd2[adj[v][i]]) cnt++;
  }
  if (cnt == (long long)adj[v].size() - 1 and
      (long long) adj[v].size() - 1 <= 2)
    dpd[v] = true;
  if (cnt == (long long)adj[v].size() - 1 and
      (long long) adj[v].size() - 1 <= 1)
    dpd2[v] = true;
}
void dfs_up(long long v, long long p) {
  long long cnt = 0;
  for (long long i = 0; i < (long long)adj[v].size(); i++) {
    if (adj[v][i] == p) continue;
    if (dpd2[adj[v][i]]) cnt++;
  }
  if (dpu2[v] and p != -1) cnt++;
  for (long long i = 0; i < (long long)adj[v].size(); i++) {
    if (adj[v][i] == p) continue;
    long long cnt2 = cnt;
    if (dpd2[adj[v][i]]) cnt2--;
    if (cnt2 == (long long)adj[v].size() - 1 and
        (long long) adj[v].size() - 1 <= 2)
      dpu[adj[v][i]] = true;
    if (cnt2 == (long long)adj[v].size() - 1 and
        (long long) adj[v].size() - 1 <= 1)
      dpu2[adj[v][i]] = true;
    dfs_up(adj[v][i], v);
  }
}
int32_t main() {
  cin >> n;
  for (long long i = 0; i < n - 1; i++)
    cin >> u >> v, u--, v--, adj[v].push_back(u), adj[u].push_back(v);
  dfs_down(0, -1), dpu2[0] = 0, dpu[0] = 0, dfs_up(0, -1);
  for (long long i = 0; i < n; i++) {
    cntt = 0;
    for (long long j = 0; j < (long long)adj[i].size(); j++) {
      if (adj[i][j] == par[i]) {
        if (!dpu[i]) cntt++;
        continue;
      }
      if (!dpd[adj[i][j]]) cntt++;
    }
    if (cntt >= 3) return cout << "No" << endl, 0;
  }
  return cout << "Yes" << endl, 0;
}
