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 = 100001;
int n;
struct Edge {
  int to, next;
} edge[N << 1];
int head[N], cnt;
void add_edge(int u, int v) {
  cnt++;
  edge[cnt].to = v;
  edge[cnt].next = head[u];
  head[u] = cnt;
  return;
}
int num[N], sum[N][6];
inline int calc(int u) {
  if (sum[u][5] || sum[u][4] || sum[u][3] > 2)
    return 5;
  else if (sum[u][3] > 1)
    return 4;
  else if (sum[u][3] || sum[u][2] || sum[u][1] > 2)
    return 3;
  else if (sum[u][1] > 1)
    return 2;
  else
    return 1;
}
void init(int u, int father) {
  for (int i = head[u]; i; i = edge[i].next) {
    int v = edge[i].to;
    if (v == father) continue;
    init(v, u);
    sum[u][num[v]]++;
  }
  num[u] = calc(u);
  return;
}
void dfs(int u, int father, int x) {
  sum[u][x]++;
  if (calc(u) < 5) {
    printf("Yes");
    exit(0);
    return;
  }
  for (int i = head[u]; i; i = edge[i].next) {
    int v = edge[i].to;
    if (v == father) continue;
    sum[u][num[v]]--;
    dfs(v, u, calc(u));
    sum[u][num[v]]++;
  }
  sum[u][x]--;
  return;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    add_edge(x, y);
    add_edge(y, x);
  }
  init(1, 0);
  dfs(1, 0, 0);
  printf("No");
  return 0;
}
