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 = 1 << 20;
int a[N];
struct quer {
  int i, l, r;
} q[N];
vector<int> ql[N], qr[N];
long long ans[N];
pair<long long, long long> T[2 * N];
pair<long long, long long> operator+=(pair<long long, long long>& a,
                                      const pair<long long, long long>& b) {
  a.first += b.first;
  a.second += b.second;
  return a;
}
inline pair<long long, long long> sum(int l, int r) {
  pair<long long, long long> ans{0, 0};
  for (l += N, r += N + 1; l < r; l >>= 1, r >>= 1) {
    if (l & 1) ans += T[l++];
    if (r & 1) ans += T[--r];
  }
  return ans;
}
inline void add(int pos, pair<long long, long long> val) {
  for (pos += N; pos; pos >>= 1) T[pos] += val;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, nq;
  cin >> n >> nq;
  for (int i = (1); i < (n + 1); ++i) cin >> a[i];
  for (int i = (0); i < (nq); ++i) q[i].i = i;
  for (int i = (0); i < (nq); ++i) {
    cin >> q[i].l;
    ql[q[i].l].push_back(i);
    ans[i] -= q[i].l;
  }
  for (int i = (0); i < (nq); ++i) {
    cin >> q[i].r;
    qr[q[i].r].push_back(i);
    ans[i] += q[i].r + 1;
  }
  a[0] = a[n + 1] = N;
  stack<int> s;
  s.push(0);
  for (int i = (1); i < (n + 1); ++i) {
    while (a[s.top()] < a[i]) s.pop();
    int lf = s.top();
    add(i, {-1, i});
    add(lf + 1, {1, -lf - 1});
    for (int qi : qr[i]) {
      pair<long long, long long> s = sum(q[qi].l, i);
      ans[qi] += q[qi].l * s.first + s.second;
    }
    s.push(i);
  }
  while (s.size()) s.pop();
  fill(T, T + 2 * N, pair<long long, long long>{0, 0});
  s.push(n + 1);
  for (int i = n; i; i--) {
    while (a[s.top()] < a[i]) s.pop();
    int rf = s.top();
    add(i, {1, -i});
    add(rf - 1, {-1, rf - 1});
    for (int qi : ql[i]) {
      pair<long long, long long> s = sum(i, q[qi].r);
      ans[qi] += q[qi].r * s.first + s.second;
    }
    s.push(i);
  }
  for (int i = (0); i < (nq); ++i) cout << ans[i] << " \n"[i == nq - 1];
}