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 = 1e6 + 10;
int n, m, a[N], top, sk[N], R[N], L[N], Lx[N], Rx[N];
struct st {
  int cn, lk, rk;
  long long ans;
} q[N];
bool cmp(st a, st b) { return a.rk < b.rk; }
bool re_cmp(st a, st b) { return a.cn < b.cn; }
long long s[N], c[N], cx[N];
vector<int> rq[N], rd[N];
int low_bit(int x) { return x & (-x); }
void add(int i) {
  int ti = i;
  if (!i) return;
  while (i <= n) c[i] += 1, cx[i] += ti, i += low_bit(i);
}
long long query(int i, long long v) {
  long long ans = 0;
  while (i > 0) ans += c[i] * v - cx[i], i -= low_bit(i);
  return ans;
}
long long tp[N];
void solve() {
  for (int i = 1, id; i <= n; i++) {
    while (rd[i].size()) add(Lx[rd[i][rd[i].size() - 1]]), rd[i].pop_back();
    while (rq[i].size()) {
      id = rq[i][rq[i].size() - 1];
      q[id].ans -= query(q[id].lk - 1, q[id].lk) - tp[q[id].lk - 1],
          rq[i].pop_back();
    }
    if (i < n) tp[i] = query(i, i + 1);
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    while (top > 0 && a[sk[top]] < a[i]) R[sk[top]] = i, --top;
    L[i] = sk[top], sk[++top] = i;
  }
  while (top > 0) R[sk[top--]] = n + 1;
  for (int i = 1; i <= m; i++) scanf("%d", &q[i].lk), q[i].cn = i;
  for (int i = 1; i <= m; i++) scanf("%d", &q[i].rk), q[i].ans = 0;
  for (int i = 1; i <= n; i++) s[i] = s[i - 1] + R[i] - L[i] - 1;
  for (int i = 1; i <= m; i++) q[i].ans = s[q[i].rk] - s[q[i].lk - 1];
  sort(q + 1, q + 1 + m, cmp);
  for (int i = 1; i <= m; i++) rq[q[i].rk].push_back(i);
  for (int i = 1; i <= n; i++)
    Lx[i] = L[i] + 1, Rx[i] = i, rd[Rx[i]].push_back(i);
  solve();
  for (int i = 1, t; i <= m; i++)
    t = q[i].rk, q[i].rk = n - q[i].lk + 1, q[i].lk = n - t + 1;
  sort(q + 1, q + 1 + m, cmp);
  for (int i = 1; i <= m; i++) rq[q[i].rk].push_back(i);
  for (int i = 1; i <= n; i++) c[i] = cx[i] = 0;
  for (int i = 1; i <= n; i++)
    Lx[i] = n - R[i] + 2, Rx[i] = n - i + 1, rd[Rx[i]].push_back(i);
  solve();
  sort(q + 1, q + 1 + m, re_cmp);
  for (int i = 1; i <= m; i++) printf("%lld ", q[i].ans);
  return 0;
}