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;
int n, q, a[1000006], la[1000006];
int l[1000006], r[1000006];
struct BIT {
  long long tr[1000006];
  inline void upd(int now, int val) {
    for (; now < 1000006; now += now & -now) tr[now] += val;
  }
  inline long long qsum(int now, long long ret = 0) {
    for (; now; now -= now & -now) ret += tr[now];
    return ret;
  }
} tr[2];
long long ans[1000006];
vector<int> G[1000006], Q[1000006];
void solve() {
  stack<int> sta;
  for (int i = 1; i <= n; i++) G[i].clear(), Q[i].clear();
  for (int i = 1; i <= q; i++) Q[l[i]].push_back(i);
  memset(tr, 0, sizeof tr);
  for (int i = 1; i <= n; i++) {
    for (; !sta.empty() && a[sta.top()] < a[i];) sta.pop();
    la[i] = !sta.empty() ? sta.top() : 0;
    sta.push(i);
    if (la[i]) {
      G[la[i]].push_back(i);
      tr[0].upd(i, i - la[i] - 1);
    } else
      tr[1].upd(i, 1), tr[0].upd(i, i);
  }
  for (int i = 1; i <= n; i++) {
    for (int u : Q[i]) {
      ans[u] += tr[0].qsum(r[u]) - tr[1].qsum(r[u]) * i;
    }
    tr[1].upd(i, -1), tr[0].upd(i, -i);
    for (int u : G[i]) tr[0].upd(u, la[u] + 1), tr[1].upd(u, 1);
  }
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= q; i++) scanf("%d", &l[i]);
  for (int i = 1; i <= q; i++) scanf("%d", &r[i]);
  solve();
  for (int i = 1; i < n - i + 1; i++) swap(a[i], a[n - i + 1]);
  for (int i = 1; i <= q; i++)
    l[i] = n - l[i] + 1, r[i] = n - r[i] + 1, swap(l[i], r[i]);
  solve();
  for (int i = 1; i <= q; i++)
    printf("%lld%c", ans[i] + r[i] - l[i] + 1, i == q ? '\n' : ' ');
}