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>
const int maxn = 1e6 + 10;
const int maxm = 1e6 + 10;
const int inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
const double eps = 1e-8;
using namespace std;
int dir[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
int n, q;
int a[maxn], l[maxn], r[maxn], li[maxn], ri[maxn];
stack<int> s;
long long ans[maxn << 2], tag[maxn << 2], res[maxn];
vector<pair<int, int> > lq[maxn], rq[maxn];
inline void push_up(int p) { ans[p] = ans[p << 1] + ans[p << 1 | 1]; }
inline void f(int p, int l, int r, long long k) {
  tag[p] += k;
  ans[p] += (r - l + 1) * k;
}
inline void push_down(int p, int l, int r) {
  int mid = (l + r) >> 1;
  f(p << 1, l, mid, tag[p]);
  f(p << 1 | 1, mid + 1, r, tag[p]);
  tag[p] = 0;
}
inline void update(int nl, int nr, int l, int r, int p, long long k) {
  if (nl <= l && r <= nr) {
    ans[p] += k * (r - l + 1);
    tag[p] += k;
    return;
  }
  push_down(p, l, r);
  int mid = (l + r) >> 1;
  if (nl <= mid) update(nl, nr, l, mid, p << 1, k);
  if (nr > mid) update(nl, nr, mid + 1, r, p << 1 | 1, k);
  push_up(p);
}
long long query(int q_x, int q_y, int l, int r, int p) {
  long long res = 0;
  if (q_x <= l && q_y >= r) return ans[p];
  int mid = (l + r) >> 1;
  push_down(p, l, r);
  if (q_x <= mid) res += query(q_x, q_y, l, mid, p << 1);
  if (q_y > mid) res += query(q_x, q_y, mid + 1, r, p << 1 | 1);
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (s.size() == 0 || a[i] <= a[s.top()])
      s.push(i);
    else {
      while (!s.empty() && a[i] > a[s.top()]) {
        int p = s.top();
        s.pop();
        ri[p] = i;
      }
      s.push(i);
    }
  }
  for (int i = 1; i <= n; i++)
    if (ri[i] == 0) ri[i] = n + 1;
  while (!s.empty()) s.pop();
  for (int i = n; i >= 1; i--) {
    if (s.size() == 0 || a[i] <= a[s.top()])
      s.push(i);
    else {
      while (!s.empty() && a[i] > a[s.top()]) {
        int p = s.top();
        s.pop();
        li[p] = i;
      }
      s.push(i);
    }
  }
  for (int i = 1; i <= q; i++) cin >> l[i];
  for (int i = 1; i <= q; i++) cin >> r[i];
  for (int i = 1; i <= q; i++) {
    lq[r[i]].push_back(pair<int, int>(l[i], i));
    rq[l[i]].push_back(pair<int, int>(r[i], i));
  }
  for (int i = 1; i <= n; i++) {
    update(li[i] + 1, i, 1, n, 1, 1);
    for (int j = 0; j < lq[i].size(); j++)
      res[lq[i][j].second] += query(lq[i][j].first, i, 1, n, 1);
  }
  memset(tag, 0, sizeof tag);
  memset(ans, 0, sizeof ans);
  for (int i = n; i >= 1; i--) {
    if (i + 1 <= ri[i] - 1) update(i + 1, ri[i] - 1, 1, n, 1, 1);
    for (int j = 0; j < rq[i].size(); j++)
      res[rq[i][j].second] += query(i, rq[i][j].first, 1, n, 1);
  }
  for (int i = 1; i <= q; i++) cout << res[i] << " ";
  return ~~(0 - 0);
}