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;
using INT = long long;
using pii = pair<int, int>;
int p[1000100], pre[1000100], l[1000100], r[1000100];
vector<pii> vec[1000100];
INT ans[1000100];
struct BIT {
  INT B[1000100];
  void init() {
    for (int i = 1; i < 1000100; i++) B[i] = 0;
  }
  void update(int u, int x) {
    for (; u < 1000100; u += u & -u) B[u] += x;
  }
  INT calc(int u, INT ans = 0) {
    for (; u; u -= u & -u) ans += B[u];
    return ans;
  }
} b[3];
void solve(int n) {
  pre[1] = 0;
  for (int i = 2; i <= n; i++) {
    if (p[i - 1] > p[i]) {
      pre[i] = i - 1;
      continue;
    }
    int pos = pre[i - 1];
    while (pos && p[pos] < p[i]) pos--;
    pre[i] = pos;
  }
  b[1].init();
  b[2].init();
  for (int i = 1; i <= n; i++) {
    if (pre[i]) {
      int x = i - pre[i] - 1;
      b[1].update(1, x);
      b[1].update(pre[i] + 1, -x);
    }
    int st = pre[i] + 1, ed = i;
    b[1].update(st, i);
    b[1].update(ed + 1, -i);
    b[2].update(st, -1);
    b[2].update(ed + 1, 1);
    for (auto p : vec[i]) {
      int id = p.second;
      int pos = p.first;
      ans[id] += b[1].calc(pos) + b[2].calc(pos) * pos;
    }
  }
}
int main() {
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  for (int i = 1; i <= q; i++) {
    scanf("%d", &l[i]);
  }
  for (int i = 1; i <= q; i++) {
    scanf("%d", &r[i]);
    vec[r[i]].push_back(pii(l[i], i));
  }
  solve(n);
  for (int i = 1; i <= n; i++) vec[i].clear();
  for (int i = 1; i <= q; i++) {
    vec[n + 1 - l[i]].push_back(pii(n + 1 - r[i], i));
  }
  reverse(p + 1, p + n + 1);
  solve(n);
  for (int i = 1; i <= q; i++) ans[i] += (r[i] - l[i] + 1);
  for (int i = 1; i <= q; i++) printf("%I64d ", ans[i]);
  printf("\n");
  return 0;
}