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;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline void wrote(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) wrote(x / 10);
  putchar(x % 10 + '0');
}
inline void writeln(long long x) {
  wrote(x);
  puts("");
}
const long long N = 200010;
vector<long long> g[N];
bool fl = 0;
long long head[N], nxt[N], vet[N], dfn[N], cur[N], n, tot, vis[N], st[N];
void insert(long long x, long long y) {
  nxt[++tot] = head[x];
  head[x] = tot;
  vet[tot] = y;
  ++dfn[x];
}
void dfs(long long x, long long pre) {
  vis[x] = 1;
  for (long long i = head[x]; i; i = nxt[i])
    if (!vis[vet[i]]) {
      if (dfn[vet[i]] >= 3) return;
      dfs(vet[i], x);
    }
}
void Dfs(long long x, long long pre) {
  long long child = 0;
  for (long long i = head[x]; i; i = nxt[i])
    if (vet[i] != pre && (dfn[vet[i]] > 0)) Dfs(vet[i], x), ++child;
  if (child > 1) {
    puts("No");
    exit(0);
  }
}
int main() {
  n = read();
  for (long long i = 2; i <= n; ++i) {
    long long x = read(), y = read();
    insert(x, y);
    insert(y, x);
  }
  for (long long i = 1; i <= n; ++i)
    if (dfn[i] == 1) dfs(i, -1);
  memcpy(st, dfn, sizeof dfn);
  memset(dfn, 0, sizeof dfn);
  for (long long x = 1; x <= n; ++x)
    if (!vis[x]) {
      for (long long i = head[x]; i; i = nxt[i])
        if (!vis[vet[i]]) g[x].push_back(vet[i]), ++dfn[vet[i]];
    }
  memcpy(cur, dfn, sizeof dfn);
  for (long long x = 1; x <= n; ++x)
    if (cur[x] == 1 && st[x] == 3) {
      dfn[x]--;
      for (long long i = 0; i < g[x].size(); i++) dfn[g[x][i]]--;
    }
  for (long long x = 1; x <= n; ++x) fl |= dfn[x] > 0;
  long long cnt = 0;
  for (long long x = 1; x <= n; ++x)
    if (dfn[x] == 1) {
      Dfs(x, -1);
      puts("Yes");
      return 0;
    }
  puts(fl ? "No" : "Yes");
}
