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 long long int N = 1e6 + 3;
const long long int LOG = 20 + 5;
const long long int MOD = 1e9 + 7;
const long long int INF = 2e9 + 10;
int n, a[N], q, lq[N], rq[N], l[N], r[N], lazy[4 * N];
long long int sum[4 * N];
vector<int> ql[N], qr[N];
stack<int> st;
long long int ans[N];
inline void shift(int b, int e, int ind) {
  if (b + 1 < e) {
    lazy[ind << 1] += lazy[ind];
    lazy[(ind << 1) | 1] += lazy[ind];
  }
  sum[ind] += 1ll * lazy[ind] * (e - b);
  lazy[ind] = 0;
  return;
}
inline void add(int l, int r, int b = 0, int e = n, int ind = 1) {
  shift(b, e, ind);
  if (r <= b || e <= l) return;
  if (l <= b && e <= r) {
    lazy[ind]++;
    shift(b, e, ind);
    return;
  }
  int mid = (b + e) >> 1;
  add(l, r, b, mid, ind << 1), add(l, r, mid, e, (ind << 1) | 1);
  sum[ind] = sum[ind << 1] + sum[(ind << 1) | 1];
  return;
}
inline long long int get(int l, int r, int b = 0, int e = n, int ind = 1) {
  shift(b, e, ind);
  if (r <= b || e <= l) return 0;
  if (l <= b && e <= r) {
    return sum[ind];
  }
  int mid = (b + e) >> 1;
  long long int res = 0;
  res = get(l, r, b, mid, ind << 1) + get(l, r, mid, e, (ind << 1) | 1);
  sum[ind] = sum[ind << 1] + sum[(ind << 1) | 1];
  return res;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < q; i++) {
    scanf("%d", &lq[i]);
    lq[i]--;
    ql[lq[i]].emplace_back(i);
  }
  for (int i = 0; i < q; i++) {
    scanf("%d", &rq[i]);
    rq[i]--;
    qr[rq[i]].emplace_back(i);
  }
  for (int i = 0; i < n; i++) {
    while (st.size() && a[st.top()] < a[i]) st.pop();
    if (st.size()) l[i] = st.top() + 1;
    st.push(i);
  }
  while (st.size()) st.pop();
  for (int i = n - 1; i >= 0; i--) {
    r[i] = n - 1;
    while (st.size() && a[st.top()] < a[i]) st.pop();
    if (st.size()) r[i] = st.top() - 1;
    st.push(i);
  }
  for (int i = 0; i < n; i++) {
    add(l[i], i + 1);
    for (int id : qr[i]) {
      ans[id] += get(lq[id], i + 1);
    }
  }
  for (int i = 0; i < 4 * N; i++) {
    sum[i] = lazy[i] = 0;
  }
  for (int i = n - 1; i >= 0; i--) {
    add(i, r[i] + 1);
    for (int id : ql[i]) {
      ans[id] += get(i, rq[id] + 1);
    }
  }
  for (int i = 0; i < q; i++) {
    printf("%lld ", ans[i] - (rq[i] - lq[i] + 1));
  }
  return 0;
}