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 INF = 1e9;
namespace T {
int size;
vector<long long> a, b;
void Init(int _n) {
  size = _n;
  a.resize(_n + 1);
  b.resize(_n + 1);
}
void Add(int x, int v) {
  for (int p = x; x <= size; x += x & -x) {
    a[x] += v;
    b[x] += p * v;
  }
}
void Modify(int l, int r) {
  Add(r, 1);
  if (l > 1) Add(l - 1, -1);
}
long long Ask(int x) {
  int p = x;
  long long r = 0;
  for (; x; x -= x & -x) {
    r += -a[x] * p + b[x];
  }
  for (x = size; x; x -= x & -x) {
    r += a[x] * p;
  }
  return r;
}
long long Query(int l, int r) { return Ask(r) - Ask(l - 1); }
}  // namespace T
struct Que {
  int x, l, r, id;
  friend bool operator<(Que a, Que b) { return a.x < b.x; }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, nq;
  cin >> n >> nq;
  vector<int> a(n + 2, INF), lf(n + 1), rf(n + 1);
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    lf[i] = i - 1;
    while (a[lf[i]] < a[i]) lf[i] = lf[lf[i]];
  }
  for (int i = n; i >= 1; --i) {
    rf[i] = i + 1;
    while (a[rf[i]] < a[i]) rf[i] = rf[rf[i]];
  }
  vector<int> ql(nq + 1), qr(nq + 1);
  vector<Que> q;
  for (int i = 1; i <= nq; ++i) {
    cin >> ql[i];
  }
  for (int i = 1; i <= nq; ++i) {
    cin >> qr[i];
    q.push_back((Que){ql[i] - 1, ql[i], qr[i], -i});
    q.push_back((Que){qr[i], ql[i], qr[i], i});
  }
  int idx = 0;
  T::Init(n);
  vector<long long> ans(nq + 1);
  sort(q.begin(), q.end());
  for (Que e : q) {
    while (idx < e.x) {
      ++idx;
      T::Modify(lf[idx] + 1, rf[idx] - 1);
    }
    if (e.id < 0) {
      ans[-e.id] -= T::Query(e.l, e.r);
    } else {
      ans[e.id] += T::Query(e.l, e.r);
    }
  }
  for (int i = 1; i <= nq; ++i) {
    cout << ans[i] << " \n"[i == nq];
  }
  return 0;
}