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;
#pragma GCC optimize("O2,no-stack-protector,unroll-loops,fast-math")
const long long maxn = 1e6 + 10, maxm = 5e3 + 10, lg = 12, mod = 1e9 + 7,
                inf = 1e18;
long long n, q, L[maxn], R[maxn], bit[maxn][2], ans[maxn], a[maxn], len[maxn];
vector<long long> qq[maxn];
void add(long long i, long long x, bool b) {
  for (i += 5; i < maxn; i += (i & -i)) bit[i][b] += x;
}
long long get(long long i, bool b) {
  long long ret = 0;
  for (i += 5; i; i -= (i & -i)) ret += bit[i][b];
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> q;
  a[n + 1] = inf;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 0; i < q; i++) cin >> L[i];
  for (int i = 0; i < q; i++) cin >> R[i], len[i] = R[i] - L[i] + 1;
  for (int T = 0; T < 2; T++) {
    memset(bit, 0, sizeof(bit));
    for (int i = 0; i < q; i++) qq[L[i]].push_back(i);
    vector<long long> vv = {n + 1};
    for (int i = n; i; i--) {
      while (a[vv.back()] < a[i]) vv.pop_back();
      long long rig = vv.back();
      vv.push_back(i);
      add(i, -i, 0);
      add(rig, rig, 0);
      add(rig, 1, 1);
      for (auto id : qq[i])
        ans[id] += get(R[id], 0) + (len[id] - get(R[id], 1)) * (R[id] + 1);
      qq[i].clear();
    }
    for (int i = 0; i < q; i++) swap(L[i] = n - L[i] + 1, R[i] = n - R[i] + 1);
    reverse(a + 1, a + 1 + n);
  }
  for (int i = 0; i < q; i++) cout << ans[i] - len[i] << ' ';
  return 0;
}