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 edge[200020], pre[200020];
int f[200020];
int deg[200020], isl[200020], leg[200020];
int n, sz;
void add(int u, int v) {
  edge[++sz] = v;
  f[sz] = pre[u];
  pre[u] = sz;
}
void dfs(int x, int fa) {
  isl[x] = 1;
  for (int p = pre[x]; p; p = f[p]) {
    int v = edge[p];
    if (v == fa) continue;
    if (deg[v] <= 2) dfs(v, x);
  }
}
int main() {
  while (~scanf("%d", &n)) {
    sz = 1;
    memset(edge, 0, sizeof(edge));
    memset(pre, 0, sizeof(pre));
    memset(f, 0, sizeof(f));
    memset(deg, 0, sizeof(deg));
    memset(isl, 0, sizeof(isl));
    memset(leg, 0, sizeof(leg));
    for (int i = 1; i < n; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      add(u, v);
      add(v, u);
      deg[u]++;
      deg[v]++;
    }
    for (int i = 1; i <= n; i++) {
      if (deg[i] == 1) dfs(i, 0);
    }
    for (int i = 1; i <= n; i++) {
      for (int p = pre[i]; p; p = f[p]) {
        int v = edge[p];
        if (isl[v]) ++leg[i];
      }
    }
    bool fuck = true;
    for (int i = 1; i <= n; i++) {
      if (isl[i]) continue;
      int ans = 0;
      for (int p = pre[i]; p; p = f[p]) {
        int v = edge[p];
        if (!isl[v] && deg[v] - min(leg[v], 2) >= 2) ++ans;
      }
      if (ans >= 3) {
        fuck = false;
        break;
      }
    }
    if (fuck)
      puts("Yes");
    else
      puts("No");
  }
  return 0;
}
