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;
long long cd, bit1[1000005], bit2[1000005], m[1000005], q, r[1000005],
    result[1000005];
long long a[1000005], b[1000005], pq[1000005];
vector<pair<long long, long long> > ruler[1000005];
vector<long long> big[1000005], tiny[1000005];
void update(long long bit[], long long x, long long val) {
  while (x <= 1000003) bit[x] += val, x += x & -x;
}
long long get(long long bit[], long long x) {
  long long k = 0;
  while (x > 0) k += bit[x], x -= x & -x;
  return k;
}
long long sum(long long bit[], long long l, long long r) {
  return get(bit, r) - get(bit, l - 1);
}
void create_r() {
  long long top = 0;
  pq[0] = cd + 1;
  m[cd + 1] = 1e9;
  for (long long i = cd; i >= 1; i--) {
    while (m[i] > m[pq[top]]) top--;
    big[pq[top]].push_back(i);
    top++, pq[top] = i;
  }
  long long pos = 1, cur = 0;
  m[0] = m[cd + 1] = 1e9;
  for (long long i = 1; i <= cd; i++) {
    for (auto v : big[i]) update(bit1, v, i - 1), update(bit2, v, 1);
    for (auto v : ruler[i]) {
      long long cur = i - v.first + 1;
      r[v.second] = sum(bit1, v.first, i) + (cur - sum(bit2, v.first, i)) * i;
    }
  }
}
void create_result() {
  long long top = 0;
  pq[0] = 0;
  m[0] = 1e9;
  for (long long i = 1; i <= cd; i++) {
    while (m[i] > m[pq[top]]) top--;
    tiny[pq[top]].push_back(i);
    top++, pq[top] = i;
  }
  for (long long i = 1; i <= cd + 1; i++) bit1[i] = bit2[i] = 0;
  long long pos = cd;
  for (long long i = cd; i >= 1; i--) {
    for (auto v : tiny[i]) update(bit1, v, i + 1), update(bit2, v, 1);
    for (auto v : ruler[i]) {
      long long cur = v.first - i + 1;
      result[v.second] = r[v.second] - sum(bit1, i, v.first) + cur -
                         (cur - sum(bit2, i, v.first)) * i;
    }
  }
}
void inp() {
  cin >> cd >> q;
  for (long long i = 1; i <= cd; i++) cin >> m[i];
  for (long long i = 1; i <= q; i++) cin >> a[i];
  for (long long i = 1; i <= q; i++) cin >> b[i];
}
void do_it() {
  for (long long i = 1; i <= q; i++)
    ruler[b[i]].push_back(pair<long long, long long>(a[i], i));
  create_r();
  for (long long i = 1; i <= cd; i++) ruler[i].clear();
  for (long long i = 1; i <= q; i++)
    ruler[a[i]].push_back(pair<long long, long long>(b[i], i));
  create_result();
  for (long long i = 1; i <= q; i++) cout << result[i] << " ";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  inp();
  do_it();
}