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 MAXN = 1e5 + 5;
int type[MAXN], pr[MAXN], cnt[MAXN];
vector<int> g[MAXN], g2[MAXN];
void dfs(int v, int p = -1) {
  pr[v] = p;
  for (int u : g[v])
    if (u != p) dfs(u, v);
  for (int u : g[v])
    if (u != p) g2[v].push_back(u);
  if ((int)((g2[v]).size()) == 0) {
    type[v] = 1;
    return;
  }
  if ((int)((g2[v]).size()) == 1 && type[g2[v][0]] == 1) {
    type[v] = 1;
    return;
  }
  if ((int)((g2[v]).size()) == 2 && type[g2[v][0]] == 1 &&
      type[g2[v][1]] == 1) {
    type[v] = 2;
    return;
  }
  type[v] = 3;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; (i) < (n - 1); ++i) {
    int v, u;
    scanf("%d%d", &v, &u);
    v--, u--;
    g[v].push_back(u), g[u].push_back(v);
  }
  dfs(0);
  for (int i = 0; (i) < (n); ++i)
    if (i != 0 && type[i] == 3) cnt[pr[i]]++;
  int v = 0;
  for (int i = 0; (i) < (n); ++i)
    if (i != 0 && cnt[i] > 1) v = i;
  for (int i = 0; (i) < (n); ++i) g2[i].clear(), cnt[i] = 0;
  dfs(v);
  for (int i = 0; (i) < (n); ++i)
    if (i != v && type[i] == 3) cnt[pr[i]]++;
  for (int i = 0; (i) < (n); ++i)
    if ((i == v && cnt[i] > 2) || (i != v && cnt[i] > 1)) {
      puts("No");
      return 0;
    }
  puts("Yes");
  return 0;
}
