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 N = 1000010;
long long tr[N], tr2[N];
void add(int p, int x, int x2) {
  for (; p < N; p += p & -p) tr[p] += x, tr2[p] += x2;
}
void add2(int l, int r, int x, int x2) {
  add(l, x, x2);
  add(r + 1, -x, -x2);
}
long long sum(int p_) {
  long long ret = 0, r2 = 0;
  for (int p = p_; p; p -= p & -p) ret += tr[p], r2 += tr2[p];
  return ret * p_ + r2;
}
int a[N], lf[N];
int stk[N], stc;
struct Node {
  int l, r, d;
} q[N];
long long ansl[N], ansr[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, qn;
  cin >> n >> qn;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= qn; i++) cin >> q[i].l;
  for (int i = 1; i <= qn; i++) cin >> q[i].r, q[i].d = i;
  for (int i = 1; i <= qn; i++) ansl[i] += q[i].r - q[i].l + 1;
  stk[0] = 1, stc = 1;
  for (int i = 1; i <= n - 1; i++) {
    while (stc && a[stk[stc - 1]] < a[i + 1]) stc--;
    if (stc) lf[i + 1] = stk[stc - 1];
    stk[stc++] = i + 1;
  }
  sort(q + 1, q + qn + 1, [](Node &a, Node &b) { return a.r < b.r; });
  int qc = 1;
  for (int i = 1; i <= n; i++) {
    if (lf[i]) add2(1, lf[i], 0, i - lf[i] - 1);
    add2(lf[i] + 1, i, -1, i);
    while (qc <= qn && q[qc].r == i) {
      ansl[q[qc].d] += sum(q[qc].l);
      qc++;
    }
  }
  memset(tr, 0, sizeof tr);
  memset(tr2, 0, sizeof tr2);
  stc = 0;
  reverse(a + 1, a + n + 1);
  for (int i = 1; i <= qn; i++)
    q[i].l = n - q[i].l + 1, q[i].r = n - q[i].r + 1, swap(q[i].l, q[i].r);
  stk[0] = 1, stc = 1;
  for (int i = 1; i <= n - 1; i++) {
    while (stc && a[stk[stc - 1]] < a[i + 1]) stc--;
    if (stc)
      lf[i + 1] = stk[stc - 1];
    else
      lf[i + 1] = 0;
    stk[stc++] = i + 1;
  }
  sort(q + 1, q + qn + 1, [](Node &a, Node &b) { return a.r < b.r; });
  qc = 1;
  for (int i = 1; i <= n; i++) {
    if (lf[i]) add2(1, lf[i], 0, i - lf[i] - 1);
    add2(lf[i] + 1, i, -1, i);
    while (qc <= qn && q[qc].r == i) {
      ansr[q[qc].d] = sum(q[qc].l);
      qc++;
    }
  }
  for (int i = 1; i <= qn; i++) cout << ansl[i] + ansr[i] << ' ';
  cout << '\n';
  return 0;
}