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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC optimize("no-stack-protector,fast-math")
using namespace std;
const long long N = 1e6 + 100, OO = 1e16 + 7, T = 1e6 + 10, M = 1e9 + 7,
                P = 6151, SQ = 280, lg = 20;
pair<long long, long long> sum[N];
long long ans[N], a[N], lf[N], l[N], r[N];
long long n, m, q[N], top;
vector<long long> ask[N], ps[N];
void upd(long long x, pair<long long, long long> add) {
  for (; x <= n; x += x & (-x))
    sum[x].first += add.first, sum[x].second += add.second;
}
pair<long long, long long> get(long long x) {
  pair<long long, long long> ans = {0, 0};
  for (; x > 0; x -= x & (-x))
    ans.first += sum[x].first, ans.second += sum[x].second;
  return ans;
}
void solve() {
  for (long long i = 1; i <= n; i++) ps[i].clear();
  top = 0;
  for (long long i = 1; i <= n; i++) {
    while (top && a[q[top]] < a[i]) top--;
    long long last = q[top];
    q[++top] = i;
    if (last)
      ps[last].push_back(i), upd(i, {i - last - 1, 0});
    else
      upd(i, {i, -1});
  }
  pair<long long, long long> tmp;
  for (long long i = 1; i <= n; i++) {
    for (auto v : ask[i]) {
      tmp = get(r[v]);
      ans[v] += tmp.first + tmp.second * i;
    }
    for (auto v : ps[i]) upd(v, {i + 1, -1});
    upd(i, {-i, 1});
  }
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i <= m; i++) cin >> l[i];
  for (long long i = 1; i <= m; i++) {
    cin >> r[i];
    ask[l[i]].push_back(i);
    ans[i] = r[i] - l[i] + 1;
  }
  solve();
  for (long long i = 1; i <= n / 2; i++) swap(a[i], a[n - i + 1]);
  for (long long i = 1; i <= n; i++) ask[i].clear();
  for (long long i = 1; i <= m; i++) {
    swap(l[i], r[i]);
    l[i] = n - l[i] + 1;
    r[i] = n - r[i] + 1;
    ask[l[i]].push_back(i);
  }
  solve();
  for (long long i = 1; i <= m; i++) cout << ans[i] << " ";
  return 0;
}