You are given an array a_1, a_2, ..., a_n and an array b_1, b_2, ..., b_n.

For one operation you can sort in non-decreasing order any subarray a[l ... r] of the array a.

For example, if a = [4, 2, 2, 1, 3, 1] and you choose subbarray a[2 ... 5], then the array turns into [4, 1, 2, 2, 3, 1]. 

You are asked to determine whether it is possible to obtain the array b by applying this operation any number of times (possibly zero) to the array a.

Input

The first line contains one integer t (1 ≤ t ≤ 3 ⋅ 10^5) — the number of queries.

The first line of each query contains one integer n (1 ≤ n ≤ 3 ⋅ 10^5).

The second line of each query contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).

The third line of each query contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ n).

It is guaranteed that ∑ n ≤ 3 ⋅ 10^5 over all queries in a test.

Output

For each query print YES (in any letter case) if it is possible to obtain an array b and NO (in any letter case) otherwise.

Example

Input


4
7
1 7 1 4 4 5 6
1 1 4 4 5 7 6
5
1 1 3 3 5
1 1 3 3 5
2
1 1
1 2
3
1 2 3
3 2 1


Output


YES
YES
NO
NO

Note

In first test case the can sort subarray a_1 ... a_5, then a will turn into [1, 1, 4, 4, 7, 5, 6], and then sort subarray a_5 ... a_6.
#include <bits/stdc++.h>
using namespace std;
using cd = complex<double>;
struct Tree {
  long long l, r, mi;
} tree[2000005];
long long a[300005], b[300005], p[300005], n;
vector<long long> c[300005], d[300005];
void buildtree(long long i, long long l, long long r) {
  tree[i].l = l;
  tree[i].r = r;
  if (tree[i].r == tree[i].l) {
    tree[i].mi = a[l];
    return;
  }
  buildtree(i * 2, l, (r + l) / 2);
  buildtree(i * 2 + 1, (l + r) / 2 + 1, r);
  tree[i].mi = min(tree[i * 2].mi, tree[i * 2 + 1].mi);
}
void update(long long i, long long l, long long r, long long k) {
  if (tree[i].l >= l && tree[i].r <= r) {
    tree[i].mi = k;
    return;
  }
  long long m = (tree[i].l + tree[i].r) / 2;
  if (l <= m) update(i * 2, l, r, k);
  if (r > m) update(i * 2 + 1, l, r, k);
  tree[i].mi = min(tree[i * 2].mi, tree[i * 2 + 1].mi);
}
long long querymin(long long i, long long l, long long r) {
  if (tree[i].l >= l && tree[i].r <= r) {
    return tree[i].mi;
  }
  long long ans = 2e9, m = (tree[i].r + tree[i].l) / 2;
  if (l <= m) ans = min(ans, querymin(i * 2, l, r));
  if (r > m) ans = min(ans, querymin(i * 2 + 1, l, r));
  return ans;
}
int main() {
  long long q;
  cin >> q;
  while (q--) {
    long long t = 0;
    cin >> n;
    for (long long i = 0; i <= n + 1; i++) {
      p[i] = 0;
      c[i].clear();
      d[i].clear();
    }
    for (long long i = 1; i <= n; i++) {
      cin >> a[i];
      c[a[i]].push_back(i);
    }
    for (long long i = 1; i <= n; i++) {
      cin >> b[i];
      d[b[i]].push_back(i);
    }
    for (long long i = 1; i <= n; i++) {
      if (c[i].size() != d[i].size()) {
        cout << "NO" << endl;
        t = 1;
        break;
      }
    }
    if (t) continue;
    buildtree(1, 1, n);
    for (long long i = 1; i <= n; i++) {
      long long x = c[b[i]][p[b[i]]];
      long long v = querymin(1, 1, x);
      p[b[i]]++;
      if (v != b[i]) {
        cout << "NO" << endl;
        t = 1;
        break;
      }
      update(1, x, x, 2e9);
    }
    if (!t) cout << "YES" << endl;
  }
}
