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, mod = 1e9 + 7;
const long long inf = 1e18;
vector<int> vl[maxn], vr[maxn];
int a[maxn], aft[maxn], bef[maxn], l[maxn], r[maxn], n;
long long lz[4 * maxn], val[4 * maxn], ans[maxn];
void Restart() {
  memset(lz, 0, sizeof lz);
  memset(val, 0, sizeof val);
}
void get(int l, int r, int id) {
  val[id] += lz[id] * (r - l);
  if (r - l > 1) {
    lz[2 * id] += lz[id];
    lz[2 * id + 1] += lz[id];
  }
  lz[id] = 0;
}
void Add(int f, int s, int x, int l = 0, int r = n, int id = 1) {
  if (f >= s || l >= r) return;
  get(l, r, id);
  if (s <= l || r <= f) return;
  if (f <= l && r <= s) {
    lz[id] += x;
    get(l, r, id);
    return;
  }
  int mid = (l + r) >> 1;
  Add(f, s, x, l, mid, 2 * id);
  Add(f, s, x, mid, r, 2 * id + 1);
  val[id] = val[2 * id] + val[2 * id + 1];
}
long long Ask(int f, int s, int l = 0, int r = n, int id = 1) {
  if (f >= s || l >= r) return 0;
  get(l, r, id);
  if (s <= l || r <= f) return 0;
  if (f <= l && r <= s) return val[id];
  int mid = (l + r) >> 1;
  return Ask(f, s, l, mid, 2 * id) + Ask(f, s, mid, r, 2 * id + 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  bef[0] = -1;
  for (int i = 1; i < n; i++) {
    bef[i] = i - 1;
    while (bef[i] >= 0 && a[bef[i]] <= a[i]) bef[i] = bef[bef[i]];
  }
  aft[n - 1] = n;
  for (int i = n - 2; i >= 0; i--) {
    aft[i] = i + 1;
    while (aft[i] < n && a[aft[i]] <= a[i]) aft[i] = aft[aft[i]];
  }
  for (int i = 0; i < q; i++) {
    cin >> l[i];
    --l[i];
    vl[l[i]].push_back(i);
  }
  for (int i = 0; i < q; i++) {
    cin >> r[i];
    --r[i];
    vr[r[i]].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    Add(bef[i] + 1, i, 1);
    for (int id : vr[i]) ans[id] += Ask(l[id], r[id] + 1);
  }
  Restart();
  for (int i = n - 1; i >= 0; i--) {
    Add(i + 1, aft[i], 1);
    for (int id : vl[i]) ans[id] += Ask(l[id], r[id] + 1);
  }
  for (int i = 0; i < q; i++) {
    cout << ans[i] + r[i] - l[i] + 1 << " ";
  }
  return cout << endl, 0;
}