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 = (int)1e6 + 10;
vector<pair<int, int>> g[N];
vector<int> v[N];
int mx[N], l[N], r[N], p[N], a[N], n, q;
long long s[N], c[N], sum[N], ans[N];
void modify(long long c[], int x, int y) {
  for (; x <= n; x += x & (-x)) c[x] += y;
}
long long get(long long c[], int x) {
  long long s = 0;
  for (; x; x -= x & (-x)) s += c[x];
  return s;
}
void modify_mx(int x, int y) {
  for (; x <= n; x += x & (-x)) mx[x] = max(mx[x], y);
}
int get_mx(int x) {
  int s = 0;
  for (; x; x -= x & (-x)) s = max(s, mx[x]);
  return s;
}
void work() {
  memset(mx, 0, sizeof(mx));
  for (int i = 1; i <= n; i++) v[i - 1].clear();
  for (int i = 1; i <= n; i++) {
    int pre = get_mx(n - p[i] + 1);
    a[i] = i - pre - 1;
    v[pre].push_back(i);
    sum[i] = sum[i - 1] + a[i];
    modify_mx(n - p[i] + 1, i);
  }
  for (int i = 1; i <= n; i++) g[i].clear();
  for (int i = 1; i <= q; i++) {
    ans[i] += sum[r[i]] - sum[l[i] - 1];
    g[l[i]].push_back(make_pair(r[i], i));
  }
  memset(s, 0, sizeof(s));
  memset(c, 0, sizeof(c));
  for (int i = 1; i <= n; i++) {
    for (const auto& j : v[i - 1]) modify(s, j, j - a[j]), modify(c, j, 1);
    for (const auto& j : g[i])
      ans[j.second] += (get(s, j.first) - get(s, i - 1)) -
                       i * (get(c, j.first) - get(c, i - 1));
  }
}
int main() {
  scanf("%d%d", &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]);
  work();
  reverse(p + 1, p + n + 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]);
  work();
  for (int i = 1; i <= q; i++)
    printf("%lld%c", ans[i] + r[i] - l[i] + 1, " \n"[i == q]);
  return 0;
}