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;
int N;
const int SIZE = 100005;
struct P {
  int next, to;
} p[2 * SIZE];
int head[SIZE];
int no;
void init() {
  memset(head, -1, sizeof(head));
  no = 0;
}
void add(int a, int b) {
  p[no].to = b;
  p[no].next = head[a];
  head[a] = no++;
}
int deg[SIZE];
int used[SIZE];
bool haveans = true;
int st = -1;
int dfs(int u) {
  used[u] = true;
  bool islef = true;
  int chd = 0;
  int cnt = 0;
  int sat = 0;
  for (int i = head[u]; i != -1; i = p[i].next) {
    int v = p[i].to;
    if (used[v]) continue;
    islef = false;
    int temp = dfs(v);
    if (temp == 1) cnt++;
    if (temp <= 2) sat++;
    chd++;
  }
  if (islef) return 1;
  if (chd - sat > 2)
    haveans = false;
  else if (chd - sat > 1 && u != st)
    haveans = false;
  if (chd == cnt)
    return cnt;
  else
    return 10000;
}
bool used1[SIZE];
int dfs1(int u) {
  used1[u] = true;
  bool islef = true;
  int chd = 0;
  int cnt = 0;
  int sat = 0;
  for (int i = head[u]; i != -1; i = p[i].next) {
    int v = p[i].to;
    if (used1[v]) continue;
    islef = false;
    int temp = dfs1(v);
    if (temp == 1) cnt++;
    if (temp <= 2) sat++;
    chd++;
  }
  if (islef) return 1;
  if (chd - sat > 1) st = u;
  if (chd == cnt)
    return cnt;
  else
    return 10000;
}
int main() {
  scanf("%d", &N);
  init();
  for (int i = 1; i < N; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    add(a, b);
    add(b, a);
    deg[a]++;
    deg[b]++;
  }
  dfs1(1);
  if (st < 0) {
    puts("Yes\n");
    return 0;
  }
  dfs(st);
  if (haveans)
    printf("Yes\n");
  else
    printf("No\n");
  return 0;
}
