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 = 100005;
int head[MAXN];
int edge[MAXN << 1];
int next1[MAXN << 1];
int sz[MAXN];
int dp[MAXN];
int vis[MAXN];
int esz;
void addedge(int u, int v) {
  edge[esz] = v;
  next1[esz] = head[u];
  head[u] = esz++;
}
void dfs(int u, int p) {
  for (int i = head[u]; i + 1; i = next1[i]) {
    int v = edge[i];
    if (v == p) continue;
    dfs(v, u);
    if (sz[v] > 2 || vis[v]) vis[u]++;
    if (vis[v] > 0 || sz[v] > 3) {
      dp[u]++;
    }
  }
}
void dfs2(int u, int p) {
  int res = vis[p];
  if (vis[u] > 0 || sz[u] > 2) res--;
  if (sz[p] > 3 || res > 0) {
    dp[u]++;
  }
  if (sz[p] > 2 || res > 0) {
    vis[u]++;
  }
  for (int i = head[u]; i + 1; i = next1[i]) {
    int v = edge[i];
    if (v == p) continue;
    dfs2(v, u);
  }
}
int main() {
  memset(head, -1, sizeof(head));
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    addedge(u, v);
    addedge(v, u);
    sz[u]++;
    sz[v]++;
  }
  for (int i = 1; i <= n; i++) {
    int res = 0;
    for (int j = head[i]; j + 1; j = next1[j]) {
      int v = edge[j];
      if (sz[v] > 3) res++;
    }
    if (res > 2) {
      printf("No\n");
      return 0;
    }
  }
  dfs(1, 0);
  dfs2(1, 0);
  for (int i = 1; i <= n; i++)
    if (dp[i] > 2) {
      printf("No\n");
      return 0;
    }
  printf("Yes\n");
  return 0;
}
