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;
vector<int> graf[100005];
int dfs(int w, int o) {
  const int JED = 0;
  const int ROZ = 1;
  const int DWA = 2;
  const int CZT = 3;
  const int OGR = 4;
  const int KON = 5;
  int licz[6] = {0, 0, 0, 0, 0, 0};
  for (int i : graf[w]) {
    if (i == o) continue;
    const int r = dfs(i, w);
    licz[r]++;
  }
  if (licz[KON]) throw -1;
  if (licz[OGR]--) {
    if (licz[OGR] or licz[CZT] or licz[DWA] or licz[ROZ] or licz[JED] > 1)
      throw -1;
    if (licz[JED]) return KON;
    return OGR;
  }
  if (licz[CZT]--) {
    if (licz[CZT] or licz[DWA] or licz[ROZ] or licz[JED] > 2) throw -1;
    if (licz[JED] == 2) return KON;
    return OGR;
  }
  if (licz[DWA]--) {
    if (licz[DWA] > 1) throw -1;
    if (licz[DWA] == 1) return CZT;
    return DWA;
  }
  if (licz[ROZ]--) return DWA;
  if (licz[JED] == 2) return ROZ;
  if (licz[JED] <= 1) return JED;
  return DWA;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    graf[a].push_back(b);
    graf[b].push_back(a);
  }
  try {
    const int r = dfs(n, -1);
    printf("Yes\n");
  } catch (int x) {
    printf("No\n");
  }
  return 0;
}
