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 = 100005;
struct Edge {
  int to, nxt;
} edge[N * 2];
bool leaf[N], ans, son[N];
int n, du[N], p1, p2, tot, fir[N];
void addedge(int x, int y) {
  du[x]++;
  du[y]++;
  edge[tot] = ((Edge){y, fir[x]});
  fir[x] = tot++;
  edge[tot] = ((Edge){x, fir[y]});
  fir[y] = tot++;
}
void dfs(int x) {
  if (leaf[x]) return;
  leaf[x] = 1;
  for (int i = fir[x]; i != -1; i = edge[i].nxt)
    if (du[edge[i].to] == 2) dfs(edge[i].to);
}
int main() {
  scanf("%d", &n);
  memset(fir, -1, sizeof(fir));
  tot = 0;
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &p1, &p2);
    addedge(p1, p2);
  }
  for (int i = 1; i <= n; i++)
    if (du[i] == 1) dfs(i);
  for (int i = 1; i <= n; i++)
    if (du[i] == 3) {
      p1 = 3;
      for (int j = fir[i]; j != -1; j = edge[j].nxt)
        if (leaf[edge[j].to]) p1--;
      if (p1 == 1) son[i] = 1;
    }
  for (int i = 1; i <= n; i++)
    if (!leaf[i])
      for (int j = fir[i]; j != -1; j = edge[j].nxt)
        if ((leaf[edge[j].to]) || (son[edge[j].to])) du[i]--;
  for (int i = 1; i <= n; i++)
    if (du[i] > 2) ans = 1;
  if (ans)
    printf("NO\n");
  else
    printf("YES\n");
  return 0;
}
