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 N = 1e6 + 10;
long long a[N], l[N], r[N], fen[N], ps[N], ans[N];
vector<long long> qry[2][N], vec;
void add(int x, int y) {
  for (x++; x < N; x += x & -x) fen[x] += y;
}
long long get(int x) {
  long long res = 0;
  for (x++; x; x -= x & -x) res += fen[x];
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  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]--;
    qry[0][l[i]].push_back(i);
    qry[1][r[i]].push_back(i);
    ans[i] = r[i] - l[i] + 1;
  }
  for (int i = 0; i < n; i++) {
    while ((int)(vec.size()) && a[vec.back()] < a[i]) {
      add(vec.back(), i - vec.back() - 1);
      vec.pop_back();
    }
    vec.push_back(i);
    ps[(int)(vec.size())] = ps[(int)(vec.size()) - 1] + i;
    for (int j : qry[1][i]) {
      long long p = lower_bound(vec.begin(), vec.end(), l[j]) - vec.begin();
      ans[j] += get(i) - get(l[j] - 1) + ((int)(vec.size()) - p) * i -
                ps[(int)(vec.size())] + ps[p];
    }
  }
  memset(fen, 0, sizeof fen);
  vec.clear();
  for (int i = n - 1; i >= 0; i--) {
    while ((int)(vec.size()) && a[-vec.back()] < a[i]) {
      add(-vec.back(), -(vec.back() + i + 1));
      vec.pop_back();
    }
    vec.push_back(-i);
    ps[(int)(vec.size())] = ps[(int)(vec.size()) - 1] + i;
    for (int j : qry[0][i]) {
      long long p = lower_bound(vec.begin(), vec.end(), -r[j]) - vec.begin();
      ans[j] += get(r[j]) - get(i - 1) -
                (((int)(vec.size()) - p) * i - ps[(int)(vec.size())] + ps[p]);
    }
  }
  for (int i = 0; i < q; i++) cout << ans[i] << ' ';
  return 0;
}