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 + 9;
long long seg[4 * N], cnt[4 * N], a[N], ans[N];
pair<long long, long long> qry[N];
vector<pair<long long, long long> > tol[N], tor[N];
void ADD(long long pos, long long l, long long r, long long ql, long long qr,
         long long val, long long is) {
  if (ql > qr || l > qr || r < ql) return;
  if (l >= ql && r <= qr) {
    seg[pos] += val;
    cnt[pos] += is;
    return;
  }
  long long mid = (l + r) / 2;
  if (is)
    ADD(pos * 2, l, mid, ql, min(qr, mid), val + max(0LL, qr - mid), is);
  else
    ADD(pos * 2, l, mid, ql, qr, val, is);
  ADD(pos * 2 + 1, mid + 1, r, ql, qr, val, is);
}
long long GET(long long pos, long long l, long long r, long long ind) {
  long long res = seg[pos] + (r - ind + 1) * cnt[pos];
  if (l == r) return res;
  long long mid = (l + r) / 2;
  if (ind <= mid)
    return res + GET(pos * 2, l, mid, ind);
  else
    return res + GET(pos * 2 + 1, mid + 1, r, ind);
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, q;
  scanf("%lld %lld", &n, &q);
  for (long long i = 0; i < n; i++) scanf("%lld", a + i);
  for (long long i = 0; i < q; i++) {
    long long x;
    scanf("%lld", &x);
    qry[i].first = x;
  }
  for (long long i = 0; i < q; i++) {
    long long x;
    scanf("%lld", &x);
    qry[i].second = x;
    qry[i].first--;
    qry[i].second--;
    tol[qry[i].second].push_back({qry[i].first, i});
    tor[n - 1 - qry[i].first].push_back({n - 1 - qry[i].second, i});
  }
  stack<long long> st;
  st.push(-1);
  for (long long i = 0; i < n; i++) {
    while ((long long)st.size() > 1 && a[st.top()] < a[i]) st.pop();
    ADD(1, 0, n - 1, st.top() + 1, i - 1, 0, 1);
    ADD(1, 0, n - 1, 0, st.top(), i - st.top() - 1, 0);
    st.push(i);
    for (pair<long long, long long> p : tol[i])
      ans[p.second] += GET(1, 0, n - 1, p.first);
  }
  reverse(a, a + n);
  fill(cnt, cnt + 4 * N, 0);
  fill(seg, seg + 4 * N, 0);
  while ((long long)st.size()) st.pop();
  st.push(-1);
  for (long long i = 0; i < n; i++) {
    while ((long long)st.size() > 1 && a[st.top()] < a[i]) st.pop();
    ADD(1, 0, n - 1, st.top() + 1, i - 1, 0, 1);
    ADD(1, 0, n - 1, 0, st.top(), i - st.top() - 1, 0);
    st.push(i);
    for (pair<long long, long long> p : tor[i])
      ans[p.second] += GET(1, 0, n - 1, p.first);
  }
  for (long long i = 0; i < q; i++)
    printf("%lld ", ans[i] + (qry[i].second - qry[i].first + 1));
  return 0;
}