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 = 1100000;
const int inf = 1e9 + 7;
int n, q;
int a[N], lt[N], rt[N], lq[N], rq[N];
long long ans[N];
struct BIT {
  long long t[N];
  void add(int x, int v) {
    for (int i = x; i <= n; i += ((i) & (-(i)))) t[i] += v;
    return;
  }
  long long sum(int x) {
    long long res = 0;
    for (int i = x; i > 0; i -= ((i) & (-(i)))) res += t[i];
    return res;
  }
} cnt, cal;
struct node {
  int l, r, id, op;
  node(int L = 0, int R = 0, int ID = 0, int OP = 0)
      : l(L), r(R), id(ID), op(OP) {}
};
vector<node> Q[N];
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  a[0] = a[n + 1] = inf;
  for (int i = 1; i <= n; i++) {
    lt[i] = i - 1;
    while (a[lt[i]] < a[i]) lt[i] = lt[lt[i]];
  }
  for (int i = n; i >= 1; i--) {
    rt[i] = i + 1;
    while (a[rt[i]] < a[i]) rt[i] = rt[rt[i]];
  }
  for (int i = 1; i <= n; i++) lt[i]++, rt[i]--;
  for (int i = 1; i <= q; i++) scanf("%d", &lq[i]);
  for (int i = 1; i <= q; i++) scanf("%d", &rq[i]);
  for (int i = 1; i <= q; i++) {
    Q[rq[i]].push_back(node(lq[i], rq[i], i, 1));
    Q[lq[i] - 1].push_back(node(lq[i], rq[i], i, -1));
    ans[i] = rq[i] - lq[i] + 1;
  }
  for (int i = 1; i <= n; i++) {
    cnt.add(lt[i], 1);
    cal.add(lt[i], lt[i]);
    for (int j = 0; j < (int)Q[i].size(); j++) {
      long long &now = ans[Q[i][j].id];
      int op = Q[i][j].op;
      int L = Q[i][j].l, R = Q[i][j].r;
      now -= 1LL * (cal.sum(R) - cal.sum(L - 1)) * op;
      now -= 1LL * (cnt.sum(L - 1)) * op * L;
    }
  }
  for (int i = 0; i <= n; i++) cnt.t[i] = cal.t[i] = 0;
  for (int i = 1; i <= n; i++) {
    cnt.add(rt[i], 1);
    cal.add(rt[i], rt[i]);
    for (int j = 0; j < (int)Q[i].size(); j++) {
      long long &now = ans[Q[i][j].id];
      int op = Q[i][j].op;
      int L = Q[i][j].l, R = Q[i][j].r;
      now += 1LL * (cal.sum(R) - cal.sum(L - 1)) * op;
      now += 1LL * (cnt.sum(n) - cnt.sum(R)) * op * R;
    }
  }
  for (int i = 1; i <= q; i++) printf("%lld ", ans[i]);
  printf("\n");
  return 0;
}