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>
const double PI = acos(-1.0);
using namespace std;
const int MAXN = 100117;
int n, a, b;
int dp[MAXN][2][3];
vector<int> g[MAXN];
int DP(int v, int x, int y, int par) {
  int &res = dp[v][x][y];
  if (res == -1) {
    int ch = ((int)(g[v]).size()) - (par != -1);
    res = 0;
    if (x == 0) {
      if (y == 0) {
        int cnt1 = 0, cnt2 = 0, cnt3 = 0;
        for (auto u : g[v])
          if (u != par) {
            if (DP(u, 0, 2, v))
              cnt2++;
            else if (DP(u, 1, 0, v))
              cnt3++;
            else if (DP(u, 0, 1, v))
              cnt1++;
          }
        if (cnt3 <= 2 && cnt1 <= 2 && cnt2 + cnt3 + cnt1 == ch) {
          res = 1;
          return res;
        }
      }
      if (y == 1) {
        int cnt1 = 0, cnt2 = 0, cnt3 = 0, cnt0 = 0;
        for (auto u : g[v])
          if (u != par) {
            if (DP(u, 1, 1, v))
              cnt0++;
            else if (DP(u, 0, 2, v))
              cnt2++;
            else if (DP(u, 1, 0, v))
              cnt3++;
            else if (DP(u, 0, 1, v))
              cnt1++;
          }
        if (cnt0 <= 1 && cnt3 <= 1 && cnt1 <= 1 && cnt2 + cnt3 + cnt1 == ch) {
          res = 1;
          return res;
        }
      }
      if (y == 2 && ch <= 2) {
        res = 1;
        for (auto u : g[v])
          if (u != par) res &= DP(u, 1, 1, v);
      }
    } else if (x == 1) {
      if (y == 0) {
        int cnt1 = 0, cnt2 = 0, cnt3 = 0;
        for (auto u : g[v])
          if (u != par) {
            if (DP(u, 0, 2, v))
              cnt2++;
            else if (DP(u, 1, 0, v))
              cnt3++;
            else if (DP(u, 0, 1, v))
              cnt1++;
          }
        if (cnt3 <= 1 && cnt1 <= 1 && cnt2 + cnt3 + cnt1 == ch) {
          res = 1;
          return res;
        }
      }
      if (y == 1 && ((int)(g[v]).size()) - (par != -1) <= 1) {
        res = 1;
        for (auto u : g[v])
          if (u != par) res &= DP(u, 1, 1, v);
      }
    }
  }
  return res;
}
int main() {
  srand(117);
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    scanf("%d %d", &a, &b);
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (int r = 0; r < 50; r++) {
    for (int i = 0; i < n; i++)
      for (int j = 0; j < 2; j++)
        for (int k = 0; k < 3; k++) dp[i][j][k] = -1;
    int v = rand() % n;
    if (DP(v, 0, 0, -1)) {
      printf("Yes\n");
      return 0;
    }
  }
  printf("No\n");
  return 0;
}
