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 = 1001000;
int n;
void add(long long* a, int x, long long y) {
  for (; x <= n; x += x & -x) a[x] += y;
}
long long gsm(long long* a, int x) {
  long long ret = 0;
  for (; x; x &= x - 1) ret += a[x];
  return ret;
}
long long sum0[N], sum1[N];
void mdf(int x, int y) {
  add(sum0, x, y);
  add(sum1, x, (long long)(1 - x) * y);
}
void modify(int l, int r, int x) {
  mdf(l, x);
  if (r < n) mdf(r + 1, -x);
}
long long query(int x) { return gsm(sum0, x) * x + gsm(sum1, x); }
vector<int> vec[N];
int q[N], l[N], r[N], a[N], m;
long long ans[N];
void solve() {
  for (int i = 1; i <= m; ++i) vec[l[i]].push_back(i);
  q[0] = n + 1;
  for (int i = n, j = 0; i; --i) {
    for (; j && a[q[j]] < a[i]; --j)
      ;
    modify(i, q[j] - 1, 1);
    q[++j] = i;
    for (int t : vec[i]) ans[t] += query(r[t]);
  }
  for (int i = 1; i <= n; ++i) vec[i].clear(), sum0[i] = sum1[i] = 0;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= m; ++i) cin >> l[i];
  for (int i = 1; i <= m; ++i) cin >> r[i];
  solve();
  for (int i = 1; i <= m; ++i)
    l[i] = n + 1 - l[i], r[i] = n + 1 - r[i], swap(l[i], r[i]);
  reverse(a + 1, a + n + 1);
  solve();
  for (int i = 1; i <= m; ++i)
    ans[i] -= r[i] - l[i] + 1, cout << ans[i] << " \n"[i == m];
  return 0;
}