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 maxn = 1e6 + 7;
long long n, a[maxn], m, Right[maxn], dem[maxn], q, dem2[maxn];
long long l[maxn], r[maxn];
stack<long long> st;
struct BIT {
  long long sad[maxn];
  void Reset() { fill_n(sad, maxn, 0); }
  void Update(long long pos, long long val) {
    for (; pos < maxn; pos += pos & -pos) {
      sad[pos] += val;
    }
  }
  long long Get(long long pos) {
    long long ans = 0;
    for (; pos > 0; pos -= (pos & -pos)) {
      ans += sad[pos];
    }
    return ans;
  }
} sum, cnt, sumid;
long long ans[maxn];
vector<long long> qr[maxn];
long long s[maxn];
void Solve() {
  sum.Reset();
  cnt.Reset();
  sumid.Reset();
  for (long long i = 1; i <= n; i++) qr[i] = {};
  while (!st.empty()) st.pop();
  for (long long i = 1; i <= q; i++) qr[r[i]].push_back(i);
  for (long long i = 1; i <= n; i++) {
    long long Left;
    while (!st.empty() && a[st.top()] < a[i]) st.pop();
    if (st.empty())
      Left = 0;
    else
      Left = st.top();
    st.push(i);
    s[i] = s[i - 1] + i - Left - 1;
    sum.Update(Left + 1, i - Left - 1);
    cnt.Update(Left + 1, 1);
    sumid.Update(Left + 1, i);
    for (auto j : qr[i]) {
      long long l = ::l[j];
      long long r = ::r[j];
      long long rest = (r - l + 1) - (cnt.Get(maxn - 1) - cnt.Get(l));
      ans[j] += sum.Get(maxn - 1) - sum.Get(l) + (r - l + 1) * (r + l) / 2 -
                sumid.Get(maxn - 1) + sumid.Get(l) - rest * l;
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (fopen("test.inp", "r"))
    freopen("test.inp", "r", stdin), freopen("test.out", "w", stdout);
  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];
  Solve();
  for (long long i = 1; i * 2 <= n; i++) swap(a[i], a[n - i + 1]);
  for (long long i = 1; i <= q; i++) {
    long long gau = l[i];
    long long meo = r[i];
    l[i] = n - meo + 1;
    r[i] = n - gau + 1;
  }
  Solve();
  for (long long i = 1; i <= q; i++) cout << ans[i] + r[i] - l[i] + 1 << ' ';
}