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;
class C_ {};
template <typename T>
C_& operator<<(C_& __m, const T& __s) {
  if (!1) cerr << "\E[91m" << __s << "\E[0m";
  return __m;
}
C_ merr;
struct __s {
  __s() {
    if (1) {
      ios_base::Init i;
      cin.sync_with_stdio(0);
      cin.tie(0);
    }
  }
  ~__s() {
    merr << "Execution time: " << (double)clock() / CLOCKS_PER_SEC << " s.\n";
  }
} __S;
vector<int> G[111111];
bool down[111111];
bool up[111111];
bool chain[111111];
void dfs(int v, int p, bool rootchain) {
  int chains = 0;
  for (int i = 0; i < (int)G[v].size(); i++) {
    int vv = G[v][i];
    if (vv == p) {
      if (rootchain) chains++;
    } else {
      dfs(vv, v, rootchain && G[v].size() < 3);
      if (chain[vv]) chains++;
    }
  }
  chain[v] = (G[v].size() == 1 ||
              (G[v].size() == 2 && chain[(G[v][0] != p ? G[v][0] : G[v][1])]));
  if (G[v].size() <= 3 && chains >= (int)G[v].size() - 1) down[v] = 1;
}
void dfs2(int v, int p) {
  int downs = 0;
  for (int i = 0; i < (int)G[v].size(); i++) {
    int vv = G[v][i];
    downs += down[vv];
    if (vv == p) continue;
    dfs2(vv, v);
  }
  up[v] = (downs >= (int)G[v].size() - 2);
}
int main(void) {
  int n;
  int root = -1;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  if (n == 1) {
    cout << "Yes" << '\n';
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (G[i].size() == 1) {
      root = i;
      break;
    }
  }
  dfs(root, -1, true);
  dfs2(root, -1);
  for (int i = 0; i < n; i++) {
    if (down[i] == 0 && up[i] == 0) {
      cout << "No" << '\n';
      return 0;
    }
  }
  cout << "Yes" << '\n';
  return 0;
}
