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 MOD = (int)1e9 + 7;
const int MAXN = (int)1e6 + 3;
const int infint = (int)1e9;
int n, q, a[MAXN], nxt[MAXN], pre[MAXN], l[MAXN], r[MAXN];
long long ans[MAXN], seg[4 * MAXN], lazy[4 * MAXN];
vector<int> v[MAXN];
stack<int> S;
void upd(int node, int st, int en, long long x) {
  seg[node] += (en - st) * x;
  lazy[node] += x;
}
void shift(int node, int st, int en) {
  int mid = (st + en) >> 1;
  upd(node << 1, st, mid, lazy[node]);
  upd(node << 1 | 1, mid, en, lazy[node]);
  lazy[node] = 0;
}
void update(int node, int st, int en, int l, int r, long long x) {
  if (l >= en || st >= r) return;
  if (l <= st && en <= r) {
    upd(node, st, en, x);
    return;
  }
  int mid = (st + en) >> 1;
  shift(node, st, en);
  update(node << 1, st, mid, l, r, x);
  update(node << 1 | 1, mid, en, l, r, x);
  seg[node] = seg[node << 1] + seg[node << 1 | 1];
}
long long get(int node, int st, int en, int l, int r) {
  if (l >= en || st >= r) return 0;
  if (l <= st && en <= r) return seg[node];
  int mid = (st + en) >> 1;
  shift(node, st, en);
  return get(node << 1, st, mid, l, r) + get(node << 1 | 1, mid, en, l, r);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = n - 1; i >= 0; i--) {
    while (!S.empty() && a[S.top()] < a[i]) S.pop();
    if (S.empty())
      nxt[i] = n;
    else
      nxt[i] = S.top();
    S.push(i);
  }
  while (!S.empty()) S.pop();
  for (int i = 0; i < n; i++) {
    while (!S.empty() && a[S.top()] < a[i]) S.pop();
    if (S.empty())
      pre[i] = -1;
    else
      pre[i] = S.top();
    S.push(i);
  }
  for (int i = 0; i < q; i++) {
    cin >> l[i];
    l[i]--;
  }
  for (int i = 0; i < q; i++) {
    cin >> r[i];
    r[i]--;
  }
  for (int i = 0; i < q; i++) v[l[i]].push_back(i);
  for (int i = n - 1; i >= 0; i--) {
    update(1, 0, n, i, nxt[i], 1);
    for (auto u : v[i]) ans[u] = get(1, 0, n, l[u], r[u] + 1);
  }
  for (int i = 0; i < n; i++) v[i].clear();
  for (int i = 0; i < 4 * n; i++) seg[i] = 0, lazy[i] = 0;
  for (int i = 0; i < q; i++) v[r[i]].push_back(i);
  for (int i = 0; i < n; i++) {
    update(1, 0, n, pre[i] + 1, i + 1, 1);
    for (auto u : v[i]) ans[u] += get(1, 0, n, l[u], r[u] + 1);
  }
  for (int i = 0; i < q; i++) cout << ans[i] - (r[i] - l[i] + 1) << " ";
}