Problem: You are given a permutation p_1, p_2, ..., p_n. You should answer q queries. Each query is a pair (l_i, r_i), and you should calculate f(l_i, r_i).

Let's denote m_{l, r} as the position of the maximum in subsegment p_l, p_{l+1}, ..., p_r.

Then f(l, r) = (r - l + 1) + f(l, m_{l,r} - 1) + f(m_{l,r} + 1, r) if l ≤ r or 0 otherwise.

Input

The first line contains two integers n and q (1 ≤ n ≤ 10^6, 1 ≤ q ≤ 10^6) — the size of the permutation p and the number of queries.

The second line contains n pairwise distinct integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, p_i ≠ p_j for i ≠ j) — permutation p.

The third line contains q integers l_1, l_2, ..., l_q — the first parts of the queries.

The fourth line contains q integers r_1, r_2, ..., r_q — the second parts of the queries.

It's guaranteed that 1 ≤ l_i ≤ r_i ≤ n for all queries.

Output

Print q integers — the values f(l_i, r_i) for the corresponding queries.

Example

Input


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


Output


1 6 8 5 1 

Note

Description of the queries: 

  1. f(2, 2) = (2 - 2 + 1) + f(2, 1) + f(3, 2) = 1 + 0 + 0 = 1; 
  2. f(1, 3) = (3 - 1 + 1) + f(1, 2) + f(4, 3) = 3 + (2 - 1 + 1) + f(1, 0) + f(2, 2) = 3 + 2 + (2 - 2 + 1) = 6; 
  3. f(1, 4) = (4 - 1 + 1) + f(1, 2) + f(4, 4) = 4 + 3 + 1 = 8; 
  4. f(2, 4) = (4 - 2 + 1) + f(2, 2) + f(4, 4) = 3 + 1 + 1 = 5; 
  5. f(1, 1) = (1 - 1 + 1) + 0 + 0 = 1. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000001;
int n;
int l[MAXN];
int r[MAXN];
int num[MAXN];
int izq[MAXN];
int der[MAXN];
long long int ans[MAXN];
long long int bit[2][MAXN];
vector<pair<int, int> > L[MAXN];
vector<pair<int, int> > R[MAXN];
void update(int id, int a, int v) {
  while (a <= n) {
    bit[id][a] += v;
    a += a & -a;
  }
}
void update_range(int a, int b, int v) {
  a++, b++;
  update(0, a, v);
  update(0, b + 1, -v);
  update(1, a, v * (a - 1));
  update(1, b + 1, -v * b);
}
long long int query(int id, int a) {
  long long int ans = 0;
  while (a > 0) {
    ans += bit[id][a];
    a -= a & -a;
  }
  return ans;
}
long long int query(int b) { return query(0, b) * b - query(1, b); }
long long int query_range(int a, int b) {
  if (a > b) return 0ll;
  a++, b++;
  return query(b) - query(a - 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int q;
  cin >> n >> q;
  vector<pair<int, int> > A;
  for (int i = 0; i < n; ++i) {
    cin >> num[i];
    A.emplace_back(num[i], i);
  }
  sort(A.begin(), A.end());
  reverse(A.begin(), A.end());
  set<int> s;
  s.insert(-1);
  s.insert(n);
  for (int i = 0; i < n; ++i) {
    auto it = s.lower_bound(A[i].second);
    der[A[i].second] = *it;
    it--;
    izq[A[i].second] = *it;
    s.insert(A[i].second);
  }
  for (int i = 0; i < q; ++i) cin >> l[i];
  for (int i = 0; i < q; ++i) cin >> r[i];
  for (int i = 0; i < q; ++i) {
    l[i]--, r[i]--;
    L[l[i]].emplace_back(r[i], i);
    R[r[i] + 1].emplace_back(l[i], i);
  }
  for (int i = n - 1; i >= 0; i--) {
    update_range(izq[i] + 1, der[i] - 1, 1);
    for (int j = 0; j < R[i].size(); ++j)
      ans[R[i][j].second] -= query_range(R[i][j].first, i - 1);
    for (int j = 0; j < L[i].size(); ++j)
      ans[L[i][j].second] += query_range(i, L[i][j].first);
  }
  for (int i = 0; i < q; ++i) cout << ans[i] << " ";
  cout << '\n';
  return 0;
}