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;
int n, Q, L[maxn], R[maxn], a[maxn];
long long ans[maxn];
pair<long long, long long> lz[maxn << 2];
vector<int> v, add1[maxn], add2[maxn];
long long tmp;
void add(int i, int j, int x, int l, int r, long long a1, long long p) {
  if (i >= j || l >= r || j <= l || i >= r) return;
  if (i >= l && j <= r) {
    lz[x].first += a1 + (i - l) * p;
    lz[x].second += p;
    return;
  }
  int md = (i + j) >> 1;
  add(i, md, x * 2, l, r, a1, p);
  add(md, j, x * 2 + 1, l, r, a1, p);
}
void get(int i, int j, int x, int l) {
  tmp += lz[x].first + lz[x].second * (l - i);
  if (i + 1 == j) return;
  int md = (i + j) >> 1;
  if (l < md)
    get(i, md, x * 2, l);
  else
    get(md, j, x * 2 + 1, l);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  scanf("%d%d", &n, &Q);
  a[0] = a[n + 1] = n + 1;
  for (int y = 1; y <= n; y++) scanf("%d", &a[y]);
  for (int y = 1; y <= Q; y++) scanf("%d", &L[y]), add2[L[y]].push_back(y);
  for (int y = 1; y <= Q; y++) scanf("%d", &R[y]), add1[R[y]].push_back(y);
  v.push_back(0);
  for (int y = 1; y <= n; y++) {
    while (a[v.back()] < a[y]) v.pop_back();
    add(1, n + 1, 1, v.back() + 1, y + 1, y - v.back() - 1, -1);
    add(1, n + 1, 1, 1, v.back() + 1, y - v.back() - 1, 0);
    v.push_back(y);
    for (auto A : add1[y]) tmp = 0, get(1, n + 1, 1, L[A]), ans[A] += tmp;
  }
  v.clear();
  v.push_back(n + 1);
  memset(lz, 0, sizeof lz);
  for (int y = n; y; y--) {
    while (a[v.back()] < a[y]) v.pop_back();
    add(1, n + 1, 1, y, v.back(), 0, 1);
    add(1, n + 1, 1, v.back(), n + 1, v.back() - 1 - y, 0);
    v.push_back(y);
    for (auto A : add2[y]) tmp = 0, get(1, n + 1, 1, R[A]), ans[A] += tmp;
  }
  for (int y = 1; y <= Q; y++) printf("%lld ", ans[y] + R[y] - L[y] + 1);
  return 0;
}