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 N = 100005;
struct Edge {
  int to, next;
} e[N << 1];
int in[N], head[N], sz;
int del[N], leg[N];
int cnt[N];
int res, n;
void addedge(int u, int v) {
  e[sz].to = v;
  e[sz].next = head[u];
  head[u] = sz++;
}
void init() {
  memset(head, -1, sizeof(head));
  memset(in, 0, sizeof(in));
  memset(del, 0, sizeof(del));
  memset(cnt, 0, sizeof(cnt));
  sz = 0;
  res = 1;
}
void dfs1(int u, int f = 0) {
  int f1 = 0, f2 = 0;
  for (int i = head[u]; i != -1; i = e[i].next) {
    int v = e[i].to;
    if (v == f) continue;
    f1 = 1;
    dfs1(v, u);
    if (del[v]) f2 = 1;
  }
  if (!f1 || (f2 && in[u] < 3)) del[u] = 1;
}
void calc(int u) {
  leg[u] = 0;
  for (int i = head[u]; i != -1; i = e[i].next) {
    int v = e[i].to;
    if (del[v]) leg[u]++;
  }
  leg[u] = min(leg[u], 2);
}
bool check(int u) {
  int cnt = 0;
  for (int i = head[u]; i != -1; i = e[i].next) {
    int v = e[i].to;
    if (!del[v] && in[v] - leg[v] > 1) ++cnt;
  }
  return cnt > 2;
}
void solve() {
  init();
  int x, y;
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    addedge(x, y);
    addedge(y, x);
    in[x]++;
    in[y]++;
  }
  int s = 0;
  for (int i = 1; i <= n; i++)
    if (in[s] < in[i]) s = i;
  if (in[s] <= 2) {
    printf("Yes\n");
    return;
  }
  dfs1(s);
  for (int i = 1; i <= n; i++) calc(i);
  int flag = 0;
  for (int i = 1; i <= n; i++) flag |= check(i);
  printf(!flag ? "Yes\n" : "No\n");
}
int main() {
  while (~scanf("%d", &n)) solve();
  return 0;
}
