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;
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
const int N = 1e6 + 5;
vector<int> v(N);
vector<pair<pair<int, int>, int>> qry;
vector<int> x(N, -1), y(N, -1);
long long segtree[4 * N][2], lazy[4 * N][2];
long long ans[N];
int n, q;
void f() {
  v[0] = N, v[n + 1] = N;
  stack<int> st;
  for (int i = 0; i <= n; i++) {
    while (!st.empty() && v[st.top()] < v[i]) st.pop();
    if (!st.empty()) x[i] = st.top();
    st.push(i);
  }
  while (!st.empty()) st.pop();
  for (int i = n + 1; i > 0; i--) {
    while (!st.empty() && v[st.top()] < v[i]) st.pop();
    if (!st.empty()) y[i] = st.top();
    st.push(i);
  }
}
bool comp_l(pair<pair<int, int>, int> &a, pair<pair<int, int>, int> &b) {
  return a < b;
}
bool comp_r(pair<pair<int, int>, int> &a, pair<pair<int, int>, int> &b) {
  return make_pair(make_pair(a.first.second, a.first.first), a.second) <
         make_pair(make_pair(b.first.second, b.first.first), b.second);
}
void spread(int node, int f, int l, int r) {
  if (lazy[node][f] != 0) {
    segtree[node][f] += 1LL * (r - l + 1) * lazy[node][f];
    if (l != r) {
      lazy[2 * node + 1][f] += lazy[node][f];
      lazy[2 * node + 2][f] += lazy[node][f];
    }
  }
  lazy[node][f] = 0;
}
void update(int x, int y, int f, int val, int node = 0, int l = 1, int r = n) {
  spread(node, f, l, r);
  if (l > y || r < x || x > y) return;
  if (x <= l && r <= y) {
    lazy[node][f] += val;
    spread(node, f, l, r);
    return;
  }
  int mid = (l + r) >> 1;
  update(x, y, f, val, 2 * node + 1, l, mid);
  update(x, y, f, val, 2 * node + 2, mid + 1, r);
  segtree[node][f] = segtree[2 * node + 1][f] + segtree[2 * node + 2][f];
}
long long query(int x, int y, int f, int node = 0, int l = 1, int r = n) {
  spread(node, f, l, r);
  if (l > y || r < x || x > y) return 0;
  if (x <= l && r <= y) return segtree[node][f];
  int mid = (l + r) >> 1;
  long long a = query(x, y, f, 2 * node + 1, l, mid);
  long long b = query(x, y, f, 2 * node + 2, mid + 1, r);
  return a + b;
}
void solve() {
  f();
  {
    sort(qry.begin(), qry.end(), comp_r);
    int idx = 1;
    for (int i = 0; i < q; i++) {
      int l = qry[i].first.first;
      int r = qry[i].first.second;
      int id = qry[i].second;
      while (idx <= r) {
        update(x[idx] + 1, idx - 1, 0, 1);
        idx++;
      }
      long long temp = query(l, r, 0);
      ans[id] += temp;
    }
  }
  {
    sort(qry.rbegin(), qry.rend(), comp_l);
    int idx = n;
    for (int i = 0; i < q; i++) {
      int l = qry[i].first.first;
      int r = qry[i].first.second;
      int id = qry[i].second;
      while (idx >= l) {
        update(idx + 1, y[idx] - 1, 1, 1);
        idx--;
      }
      long long temp = query(l, r, 1);
      ans[id] += temp;
    }
  }
  for (int i = 0; i < q; i++) {
    int l = qry[i].first.first;
    int r = qry[i].first.second;
    int idx = qry[i].second;
    ans[idx] += (r - l + 1);
  }
  for (int i = 0; i < q; i++) cout << ans[i] << ' ';
  cout << "\n";
}
int main() {
  fast();
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> v[i];
  for (int i = 0; i < q; i++) {
    int a;
    cin >> a;
    qry.push_back(make_pair(make_pair(a, -1), i));
  }
  for (int i = 0; i < q; i++) cin >> qry[i].first.second;
  solve();
}