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;
template <typename TT>
inline void read(TT &x) {
  x = 0;
  bool f = false;
  char in = getchar();
  while (!isdigit(in)) {
    if (in == '-') f = true;
    in = getchar();
  }
  while (isdigit(in)) x = x * 10 + in - '0', in = getchar();
  x = f ? -x : x;
}
const int N = 1e6 + 9;
struct Ask {
  int l, r;
};
int n, m;
int a[N], lb[N], rb[N], stk[N];
long long ans[N];
Ask q[N];
vector<int> qry[N], del[N];
struct Bit {
  long long tr[N];
  inline void clear() { memset(tr, 0, sizeof tr); }
  inline void update(int x, long long k) {
    for (int i = x; i <= n; i += i & -i) tr[i] += k;
  }
  inline long long query(int x) {
    long long re = 0;
    for (int i = x; i; i -= i & -i) re += tr[i];
    return re;
  }
} tr1, tr2;
int main() {
  read(n), read(m);
  for (int i = 1; i <= n; ++i) read(a[i]);
  for (int i = 1; i <= m; ++i) read(q[i].l);
  for (int i = 1; i <= m; ++i) read(q[i].r);
  int top = 0;
  for (int i = 1; i <= n; ++i) {
    while (top && a[stk[top]] < a[i]) rb[stk[top--]] = i;
    if (top) lb[i] = stk[top];
    stk[++top] = i;
  }
  while (top) rb[stk[top--]] = n + 1;
  for (int i = 1; i <= n; ++i) del[lb[i]].push_back(i);
  for (int i = 1; i <= m; ++i) qry[q[i].l].push_back(i);
  for (int i = 1; i <= n; ++i) tr1.update(i, lb[i]);
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < del[i - 1].size(); ++j) {
      tr1.update(del[i - 1][j], -lb[del[i - 1][j]]);
      tr2.update(del[i - 1][j], 1);
    }
    for (int j = 0; j < qry[i].size(); ++j) {
      int id = qry[i][j];
      ans[id] -= tr1.query(q[id].r) - tr1.query(q[id].l - 1);
      ans[id] -=
          (long long)(i - 1) * (tr2.query(q[id].r) - tr2.query(q[id].l - 1));
    }
  }
  for (int i = 0; i <= n; ++i) del[i].clear(), qry[i].clear();
  for (int i = 1; i <= n; ++i) del[rb[i]].push_back(i);
  for (int i = 1; i <= m; ++i) qry[q[i].r].push_back(i);
  tr1.clear(), tr2.clear();
  for (int i = 1; i <= n; ++i) tr1.update(i, rb[i] - 1);
  for (int i = n; i; --i) {
    for (int j = 0; j < del[i + 1].size(); ++j) {
      tr1.update(del[i + 1][j], -rb[del[i + 1][j]] + 1);
      tr2.update(del[i + 1][j], 1);
    }
    for (int j = 0; j < qry[i].size(); ++j) {
      int id = qry[i][j];
      ans[id] += tr1.query(q[id].r) - tr1.query(q[id].l - 1);
      ans[id] += (long long)i * (tr2.query(q[id].r) - tr2.query(q[id].l - 1));
    }
  }
  for (int i = 1; i <= m; ++i) printf("%lld ", ans[i]);
  puts("");
  return 0;
}