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 MAX_N = 1000 * 1000 + 5;
int n, q, p[MAX_N], l[MAX_N], r[MAX_N], last[MAX_N], lazy[4 * MAX_N];
long long seg[4 * MAX_N], ans[MAX_N];
vector<int> qr[MAX_N];
void read_input() {
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> p[i], p[i]--;
  p[n] = n;
  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]--;
    qr[l[i]].push_back(i);
  }
}
void find_last() {
  vector<int> st = {n};
  for (int i = n - 1; ~i; i--) {
    while (p[st.back()] < p[i]) st.pop_back();
    last[i] = st.back();
    st.push_back(i);
  }
}
void update(int id, int l, int r) {
  seg[id] += 1LL * (r - l) * lazy[id];
  if (l < r - 1) {
    lazy[id << 1] += lazy[id];
    lazy[id << 1 | 1] += lazy[id];
  }
  lazy[id] = 0;
}
void add(int x, int y, int id = 1, int l = 0, int r = n) {
  update(id, l, r);
  if (x >= r || y <= l) return;
  if (x <= l && y >= r) {
    lazy[id]++;
    update(id, l, r);
    return;
  }
  int mid = (l + r) >> 1;
  add(x, y, id << 1, l, mid);
  add(x, y, id << 1 | 1, mid, r);
  seg[id] = seg[id << 1] + seg[id << 1 | 1];
}
long long get(int x, int y, int id = 1, int l = 0, int r = n) {
  update(id, l, r);
  if (x >= r || y <= l) return 0;
  if (x <= l && y >= r) return seg[id];
  int mid = (l + r) >> 1;
  return get(x, y, id << 1, l, mid) + get(x, y, id << 1 | 1, mid, r);
}
void update_ans(bool e = false) {
  for (int i = n - 1; ~i; i--) {
    add(i, last[i]);
    for (int j : qr[i]) {
      ans[j] += get(0, r[j] + 1);
      if (e) ans[j] -= r[j] - i + 1;
    }
  }
}
void solve() {
  find_last();
  update_ans();
  reverse(p, p + n);
  memset(seg, 0, sizeof seg);
  memset(lazy, 0, sizeof lazy);
  for (int i = 0; i < n; i++) qr[i].clear();
  for (int i = 0; i < q; i++) {
    int r_tmp = n - 1 - l[i], l_tmp = n - 1 - r[i];
    l[i] = l_tmp;
    r[i] = r_tmp;
    qr[l[i]].push_back(i);
  }
  find_last();
  update_ans(true);
}
void print_output() {
  for (int i = 0; i < q; i++) cout << ans[i] << ' ';
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  read_input();
  solve();
  print_output();
}