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 MOD = 1000000007;
const int INF = 2000000000;
const long long BIG = 1446803456761533460LL;
const int maxn = 1000010;
int n, q, A[maxn], lt[maxn], rt[maxn], L[maxn], R[maxn];
long long res[maxn];
vector<pair<int, int> > ql[maxn], qr[maxn];
struct Fenwick {
  long long fen[maxn];
  void update(int i, int v) {
    for (++i; i < maxn; i += i & -i) fen[i] += v;
  }
  long long get(int i) {
    long long s = 0;
    for (++i; i; i &= i - 1) s += fen[i];
    return s;
  }
  void clear() { memset(fen, 0, sizeof fen); }
} fen1, fen2;
void add(int l, int r) {
  fen1.update(l, -l);
  fen2.update(l, 1);
  fen1.update(r, r);
  fen2.update(r, -1);
}
long long sum(int l, int r) {
  return fen1.get(r) - fen1.get(l) + fen2.get(r) * r - fen2.get(l) * l;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n; ++i) cin >> A[i];
  for (int i = 0; i < n; ++i) {
    for (lt[i] = i - 1; lt[i] >= 0 && A[lt[i]] < A[i]; lt[i] = lt[lt[i]])
      ;
  }
  for (int i = (n)-1; i >= 0; --i) {
    for (rt[i] = i + 1; rt[i] < n && A[rt[i]] < A[i]; rt[i] = rt[rt[i]])
      ;
  }
  for (int i = 0; i < q; ++i) cin >> L[i], --L[i];
  for (int i = 0; i < q; ++i) cin >> R[i], --R[i];
  for (int i = 0; i < q; ++i) {
    ql[L[i]].emplace_back(R[i], i);
    qr[R[i]].emplace_back(L[i], i);
  }
  for (int i = 0; i < n; ++i) {
    add(lt[i] + 1, i);
    for (auto [l, qi] : qr[i]) res[qi] += sum(l, i + 1);
  }
  fen1.clear(), fen2.clear();
  for (int i = (n)-1; i >= 0; --i) {
    add(i, rt[i]);
    for (auto [r, qi] : ql[i]) res[qi] += sum(i, r + 1);
  }
  for (int i = 0; i < q; ++i) cout << res[i] << ' ';
  cout << '\n';
}