Gildong was hiking a mountain, walking by millions of trees. Inspired by them, he suddenly came up with an interesting idea for trees in data structures: What if we add another edge in a tree?

Then he found that such tree-like graphs are called 1-trees. Since Gildong was bored of solving too many tree problems, he wanted to see if similar techniques in trees can be used in 1-trees as well. Instead of solving it by himself, he's going to test you by providing queries on 1-trees.

First, he'll provide you a tree (not 1-tree) with n vertices, then he will ask you q queries. Each query contains 5 integers: x, y, a, b, and k. This means you're asked to determine if there exists a path from vertex a to b that contains exactly k edges after adding a bidirectional edge between vertices x and y. A path can contain the same vertices and same edges multiple times. All queries are independent of each other; i.e. the added edge in a query is removed in the next query.

Input

The first line contains an integer n (3 ≤ n ≤ 10^5), the number of vertices of the tree.

Next n-1 lines contain two integers u and v (1 ≤ u,v ≤ n, u ≠ v) each, which means there is an edge between vertex u and v. All edges are bidirectional and distinct.

Next line contains an integer q (1 ≤ q ≤ 10^5), the number of queries Gildong wants to ask.

Next q lines contain five integers x, y, a, b, and k each (1 ≤ x,y,a,b ≤ n, x ≠ y, 1 ≤ k ≤ 10^9) – the integers explained in the description. It is guaranteed that the edge between x and y does not exist in the original tree.

Output

For each query, print "YES" if there exists a path that contains exactly k edges from vertex a to b after adding an edge between vertices x and y. Otherwise, print "NO".

You can print each letter in any case (upper or lower).

Example

Input


5
1 2
2 3
3 4
4 5
5
1 3 1 2 2
1 4 1 3 2
1 4 1 3 3
4 2 3 3 9
5 2 3 3 9


Output


YES
YES
NO
YES
NO

Note

The image below describes the tree (circles and solid lines) and the added edges for each query (dotted lines).

<image>

Possible paths for the queries with "YES" answers are: 

  * 1-st query: 1 – 3 – 2 
  * 2-nd query: 1 – 2 – 3 
  * 4-th query: 3 – 4 – 2 – 3 – 4 – 2 – 3 – 4 – 2 – 3 
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using Pll = pair<ll, ll>;
using Pii = pair<int, int>;
constexpr int INF = 1 << 30;
constexpr ll LINF = 1LL << 60;
constexpr ll MOD = 1000000007;
constexpr long double EPS = 1e-10;
constexpr int dyx[4][2] = {{0, 1}, {-1, 0}, {0, -1}, {1, 0}};
constexpr int N_MAX = 100000;
vector<int> graph[N_MAX];
class LowestCommonAncestor {
 public:
  int root, n, LOG_V_MAX = 30;
  vector<int> depth;
  vector<vector<int>> parent;
  LowestCommonAncestor(int root = 0, int n = N_MAX) : root(root), n(n) {
    depth.assign(n, -1);
    parent.assign(n, vector<int>(LOG_V_MAX, -1));
    dfs(root, -1, 0);
    for (int j = 1; j < LOG_V_MAX; ++j) {
      for (int i = 0; i < n; ++i) {
        if (parent[i][j - 1] == -1) continue;
        parent[i][j] = parent[parent[i][j - 1]][j - 1];
      }
    }
  }
  void dfs(int node, int par, int d) {
    depth[node] = d;
    parent[node][0] = par;
    for (auto child : graph[node]) {
      int c = child;
      if (c == par) continue;
      dfs(c, node, d + 1);
    }
  }
  int get_lca(int u, int v) {
    if (depth[u] > depth[v]) swap(u, v);
    int depth_diff = depth[v] - depth[u];
    for (int j = 0; j < LOG_V_MAX; ++j) {
      if (depth_diff & (1 << j)) {
        v = parent[v][j];
      }
    }
    if (u == v) return u;
    for (int j = LOG_V_MAX - 1; j >= 0; --j) {
      if (parent[u][j] != parent[v][j]) {
        u = parent[u][j];
        v = parent[v][j];
      }
    }
    return parent[u][0];
  }
  int get_dist(int u, int v) {
    return depth[u] + depth[v] - 2 * depth[get_lca(u, v)];
  }
};
bool in_loop(LowestCommonAncestor &lca, int a, int x, int y, int l) {
  if (lca.depth[a] < lca.depth[l] ||
      max(lca.depth[x], lca.depth[y]) < lca.depth[a])
    return false;
  if (lca.get_lca(a, l) == l) {
    if (lca.get_lca(a, x) == a || lca.get_lca(a, y) == a) return true;
  }
  return false;
}
void solve(LowestCommonAncestor &lca) {
  int x, y, a, b, k;
  cin >> x >> y >> a >> b >> k;
  --x;
  --y;
  --a;
  --b;
  int xy = lca.get_dist(x, y), ab = lca.get_dist(a, b);
  if (ab <= k && k % 2 == ab % 2) {
    cout << "YES\n";
    return;
  }
  int l = lca.get_lca(x, y);
  int cycle_length = xy + 1;
  int d = lca.get_dist(a, x) + 1 + lca.get_dist(y, b);
  if (d <= k && d % 2 == k % 2) {
    cout << "YES\n";
    return;
  }
  d = d - 1 + cycle_length - xy;
  if (d <= k && d % 2 == k % 2) {
    cout << "YES\n";
    return;
  }
  d = lca.get_dist(a, y) + 1 + lca.get_dist(x, b);
  if (d <= k && d % 2 == k % 2) {
    cout << "YES\n";
    return;
  }
  d = d - 1 + cycle_length - xy;
  if (d <= k && d % 2 == k % 2) {
    cout << "YES\n";
    return;
  }
  cout << "NO\n";
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  LowestCommonAncestor lca = LowestCommonAncestor();
  int t;
  cin >> t;
  while (t--) {
    solve(lca);
  }
}
