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 long long N = 1e6 + 5;
vector<pair<long long, long long>> vec[N];
long long n;
long long tr1[N], tr2[N];
void upt(long long x, long long c) {
  long long p = x;
  while (x <= n) tr1[x] += c, tr2[x] += p * c, x += (x & -x);
}
void add(long long l, long long r, long long c) { upt(l, c), upt(r + 1, -c); }
long long query(long long x) {
  long long res = 0, p = x;
  while (x > 0) res += (p + 1) * tr1[x], res -= tr2[x], x -= (x & -x);
  return res;
}
long long solve(long long l, long long r) { return query(r) - query(l - 1); }
long long L[N], R[N], a[N], st[N], top;
long long ans[N];
void calc() {
  memset(tr1, 0, sizeof(tr1));
  memset(tr2, 0, sizeof(tr2));
  top = 0;
  for (long long i = 1; i <= n; ++i) {
    long long now = top;
    while (top >= 1 && a[st[top]] < a[i]) --top;
    if (top)
      add(i, i, solve(st[top], st[top]) + 1);
    else
      add(i, i, 1);
    if (now != top) add(st[top] + 1, i - 1, 1);
    st[++top] = i;
    for (auto p : vec[i]) {
      long long x = *lower_bound(st + 1, st + top + 1, p.first);
      ans[p.second] += solve(x + 1, i) - 1ll * (i - x) * solve(x, x);
    }
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  long long q;
  cin >> n >> q;
  for (long long i = 1; i <= n; ++i) cin >> a[i];
  for (long long i = 1; i <= q; ++i) cin >> L[i];
  for (long long i = 1; i <= q; ++i) cin >> R[i];
  for (long long i = 1; i <= q; ++i)
    vec[R[i]].push_back({L[i], i}), ans[i] = R[i] - L[i] + 1;
  calc();
  for (long long i = 1; i <= n; ++i) vec[i].clear();
  reverse(a + 1, a + n + 1);
  for (long long i = 1; i <= q; ++i) {
    swap(L[i], R[i]);
    L[i] = n - L[i] + 1, R[i] = n - R[i] + 1;
    vec[R[i]].push_back({L[i], i});
  }
  calc();
  for (long long i = 1; i <= q; ++i) cout << ans[i] << " ";
}