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 maxn = 1e6 + 20;
vector<int> query[maxn];
int l[maxn], r[maxn], a[maxn], n;
long long ans[maxn], fen[2][maxn];
void add(int p, int val, int id) {
  for (p += 5; p < maxn; p += ((p) & (-(p)))) fen[id][p] += val;
}
long long get(int p, int id) {
  long long res = 0;
  for (p += 5; p > 0; p -= ((p) & (-(p)))) res += fen[id][p];
  return res;
}
int main() {
  int q;
  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", &l[i]);
    l[i]--;
  }
  for (int i = 0; i < q; i++) {
    scanf("%d", &r[i]);
    r[i]--;
  }
  for (int _ = 0; _ < 2; _++) {
    memset(fen, 0, sizeof fen);
    for (int i = 0; i < q; i++) query[l[i]].push_back(i);
    stack<int> st;
    for (int i = n - 1; i >= 0; i--) {
      while (!st.empty() && a[st.top()] < a[i]) st.pop();
      int pos = n;
      if (!st.empty()) pos = st.top();
      st.push(i);
      add(i, -(i - 1), 0);
      add(pos, i - 1, 0);
      add(i, 1, 1);
      add(pos, -1, 1);
      add(pos, pos - i, 0);
      for (auto ind : query[i]) {
        ans[ind] += get(r[ind], 0);
        ans[ind] += 1LL * r[ind] * get(r[ind], 1);
      }
      query[i].clear();
    }
    for (int i = 0; i < q; i++) {
      l[i] = n - l[i] - 1;
      r[i] = n - r[i] - 1;
      swap(l[i], r[i]);
    }
    reverse(a, a + n);
  }
  for (int i = 0; i < q; i++) printf("%lld ", ans[i] - r[i] + l[i] - 1);
  printf("\n");
}