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;
vector<int> con[100010];
int dap, cnt2[100010], n, x, y, ch[100010], cnt[100010], sw;
void dfs(int x, int bf) {
  int i;
  for (i = 0; i < con[x].size(); i++) {
    if (bf == con[x][i]) continue;
    cnt[x]++;
    dfs(con[x][i], x);
  }
  if (cnt[x] == 0) ch[x] = 1;
  sw = 0;
  for (i = 0; i < con[x].size(); i++) {
    if (bf == con[x][i]) continue;
    if (cnt[x] == 1 && ch[con[x][i]] == 1) ch[x] = 1;
    if (cnt[x] == 2 && ch[con[x][i]] == 1) sw++;
    if (ch[con[x][i]] != 1) {
      ch[x] = 0;
      return;
    }
  }
  if (cnt[x] == 2 && sw == 2) ch[x] = 2;
  if (cnt[x] >= 3) ch[x] = 0;
}
void dfs2(int x, int bf) {
  int i;
  for (i = 0; i < con[x].size(); i++) {
    if (bf == con[x][i]) continue;
    if (ch[con[x][i]] == 0) cnt2[x]++;
    dfs2(con[x][i], x);
  }
}
void dfs3(int x, int bf, int par) {
  int i, c = 0, p = 0;
  if (x != 1) c++;
  if (par >= 2) {
    if (par >= 3) cnt2[x]++;
    for (i = 0; i < con[x].size(); i++) {
      if (bf == con[x][i]) continue;
      dfs3(con[x][i], x, 3);
    }
    return;
  }
  for (i = 0; i < con[x].size(); i++) {
    if (bf == con[x][i]) continue;
    if (ch[con[x][i]] == 1) c++;
    if (ch[con[x][i]] != 1) {
      if (p) c = 1e9;
      p = i + 1;
    }
  }
  if (c >= 1e9) {
    for (i = 0; i < con[x].size(); i++) {
      if (bf == con[x][i]) continue;
      if (ch[con[x][i]] != 1) dfs3(con[x][i], x, 3);
    }
    return;
  }
  if (p) {
    dfs3(con[x][p - 1], x, c);
    return;
  }
  for (i = 0; i < con[x].size(); i++) {
    if (bf == con[x][i]) continue;
    dfs3(con[x][i], x, c - ch[con[x][i]]);
  }
}
int main() {
  int i;
  scanf("%d", &n);
  for (i = 1; i < n; i++) {
    scanf("%d %d", &x, &y);
    con[x].push_back(y);
    con[y].push_back(x);
  }
  dfs(1, 0);
  dfs2(1, 0);
  dfs3(1, 0, 0);
  for (i = 1; i <= n; i++) {
    if (cnt2[i] >= 3) dap = 1;
  }
  if (dap == 1)
    printf("No");
  else
    printf("Yes");
  return 0;
}
