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 int MXN = 128 * 1024;
int n;
int e[MXN][2];
int ne[MXN];
vector<vector<int> > G;
static int dfs(int p, int v) {
  int nc[5] = {0};
  for (size_t i = 0; i < G[v].size(); i++) {
    int tg = G[v][i];
    if (tg == p) continue;
    int k = dfs(v, tg);
    if (k < 0) return -1;
    if (k == 5) return -1;
    nc[k]++;
  }
  {
    int pr[5];
    pr[0] = nc[0];
    for (int i = 1; i < 5; i++) pr[i] = pr[i - 1] + nc[i];
    if (nc[4]) {
      if (pr[4] == 1) return 4;
      if (pr[4] > 2) return -1;
      if (pr[4] - pr[0] > 1) return -1;
      return 5;
    }
    if (nc[3]) {
      if (pr[4] - pr[0] > 1) return -1;
      if (nc[0] > 2) return -1;
      if (nc[0] == 2) return 5;
      return 4;
    }
    if (nc[2]) {
      if (nc[2] > 2) return -1;
      if (nc[2] == 2) {
        return 3;
      } else {
        return 2;
      }
    }
    if (nc[1]) return 2;
    if (nc[0] > 2) return 2;
    if (nc[0] == 2) return 1;
    return 0;
  }
}
static bool solve() {
  G.clear();
  G.resize(n);
  memset(ne, 0, sizeof(ne));
  for (int i = 0; i < n - 1; i++) {
    ne[e[i][0]]++;
    ne[e[i][1]]++;
  }
  for (int i = 0; i < n; i++) G[i].reserve(ne[i]);
  for (int i = 0; i < n - 1; i++) {
    G[e[i][0]].push_back(e[i][1]);
    G[e[i][1]].push_back(e[i][0]);
  }
  int k = dfs(0, 0);
  return k >= 0;
}
int main() {
  while (scanf("%d", &n) == 1) {
    for (int i = 0; i < n - 1; i++) {
      scanf("%d %d", &e[i][0], &e[i][1]);
      e[i][0]--;
      e[i][1]--;
    }
    puts(solve() ? "Yes" : "No");
  }
  return 0;
}
