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 + 10;
pair<int, long long> C[maxn];
long long ans[maxn];
vector<pair<int, int> > Ql[maxn];
priority_queue<pair<int, int> > heap;
int n, m, da[maxn], stk[maxn], LMax[maxn];
struct Que {
  int l, r;
} q[maxn];
void update(int x, int add) {
  for (int i = x; i <= n; i += ((i) & (-i))) ++C[i].first, C[i].second += add;
}
pair<int, long long> query(int x) {
  pair<int, long long> ret = make_pair(0, 0ll);
  for (int i = x; i; i -= ((i) & (-i))) {
    ret.first += C[i].first;
    ret.second += C[i].second;
  }
  return ret;
}
void Solve() {
  int top = 0;
  while (!heap.empty()) heap.pop();
  for (int i = 1; i <= n; ++i) LMax[i] = 0, Ql[i].clear();
  for (int i = n; i >= 1; --i) {
    while (top && da[stk[top]] < da[i]) LMax[stk[top--]] = i;
    stk[++top] = i;
  }
  for (int i = 1; i <= n; ++i) heap.push(make_pair(LMax[i] + 1, i));
  for (int i = 1; i <= m; ++i) Ql[q[i].l].push_back(make_pair(q[i].r, i));
  heap.push(make_pair(-1, -1));
  for (int i = 1; i <= n; ++i) C[i] = make_pair(0, 0ll);
  for (int i = n; i >= 1; --i) {
    pair<int, int> u = heap.top();
    while (u.first >= i) {
      update(u.second, u.first);
      heap.pop();
      u = heap.top();
    }
    for (int k = 0, j, num; k < (int)Ql[i].size(); ++k) {
      j = Ql[i][k].first;
      pair<int, long long> qi = query(i - 1), qj = query(j);
      num = j - i + 1 - (qj.first - qi.first);
      ans[Ql[i][k].second] += 1ll * (i + j) * (j - i + 1) / 2 -
                              (qj.second - qi.second + 1ll * i * num);
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &da[i]);
  for (int i = 1; i <= m; ++i) scanf("%d", &q[i].l);
  for (int i = 1; i <= m; ++i) {
    scanf("%d", &q[i].r);
    ans[i] = q[i].r - q[i].l + 1;
  }
  Solve();
  reverse(da + 1, da + n + 1);
  for (int i = 1; i <= m; ++i) {
    q[i].l = n - q[i].l + 1;
    q[i].r = n - q[i].r + 1;
    swap(q[i].l, q[i].r);
  }
  Solve();
  for (int i = 1; i <= m; ++i) printf("%I64d ", ans[i]);
  return 0;
}