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>
const int MAX_N = 100000;
int n, nodes[MAX_N + 10], en, nxt[MAX_N * 2 + 10], to[MAX_N * 2 + 10];
int deg[MAX_N + 10];
bool ok[MAX_N + 10];
int dep[MAX_N + 10], cnt;
inline void addEdge(int f, int t) {
  ++en;
  to[en] = t;
  nxt[en] = nodes[f];
  nodes[f] = en;
}
bool dfs2(int cur, int fa) {
  if (deg[cur] > 2) --cnt;
  for (int e = nodes[cur]; e; e = nxt[e]) {
    if (to[e] == fa) continue;
    if (deg[to[e]] == 3) --cnt;
  }
  if (cnt == 0) return true;
  for (int e = nodes[cur]; e; e = nxt[e]) {
    if (to[e] == fa) continue;
    if (deg[to[e]] == 3) ++cnt;
    if (dfs2(to[e], cur)) return true;
    if (deg[to[e]] == 3) --cnt;
  }
  for (int e = nodes[cur]; e; e = nxt[e]) {
    if (to[e] == fa) continue;
    if (deg[to[e]] == 3) ++cnt;
  }
  if (deg[cur] > 2) ++cnt;
  return false;
}
void dfs(int cur, int fa) {
  for (int e = nodes[cur]; e; e = nxt[e]) {
    if (to[e] == fa) continue;
    dep[to[e]] = dep[cur] + 1;
    dfs(to[e], cur);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    addEdge(a, b);
    addEdge(b, a);
    ++deg[a];
    ++deg[b];
  }
  dfs(1, 0);
  int g = 0;
  for (int i = 1; i <= n; i++)
    if (deg[i] > 2) {
      ++cnt;
      if (dep[i] > dep[g]) g = i;
    }
  if (!g)
    printf("Yes\n");
  else {
    if (dfs2(g, 0))
      printf("Yes\n");
    else
      printf("No\n");
  }
  return 0;
}
