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 MAXN = 1e6 + 9;
int n, m;
int w[MAXN], ql[MAXN], qr[MAXN], p[MAXN], bit1[MAXN], val[MAXN];
long long ans[MAXN], temp[MAXN], bit2[MAXN];
vector<int> g[MAXN];
vector<pair<int, int>> q[MAXN];
template <class T>
void add(int x, int v, T b[]) {
  for (; x <= n; x += x & -x) b[x] += v;
}
template <class T>
long long que(int x, T b[]) {
  long long ret = 0;
  for (; x; x -= x & -x) ret += b[x];
  return ret;
}
void solve(long long res[]) {
  for (int i = 1; i <= n; i++) {
    g[i].clear();
    q[i].clear();
    bit1[i] = 0;
    bit2[i] = 0;
  }
  w[0] = MAXN;
  p[0] = 0;
  vector<int> stk(1, 0);
  for (int i = 1; i <= n; i++) {
    while (w[stk.back()] < w[i]) stk.pop_back();
    p[i] = stk.back() + 1;
    stk.emplace_back(i);
    g[p[i]].emplace_back(i);
    add(i, val[p[i]], bit2);
  }
  for (int i = 0; i < m; i++) q[ql[i]].emplace_back(qr[i], i);
  for (int i = 1; i <= n; i++) {
    for (auto e : q[i])
      res[e.second] = val[i] * (que(e.first, bit1) - que(i - 1, bit1)) +
                      que(e.first, bit2) - que(i - 1, bit2);
    for (auto e : g[i]) {
      add(e, 1, bit1);
      add(e, -val[p[e]], bit2);
    }
  }
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
  for (int i = 0; i < m; i++) scanf("%d", &ql[i]);
  for (int i = 0; i < m; i++) scanf("%d", &qr[i]);
  for (int i = 1; i <= n; i++) val[i] = i;
  solve(temp);
  for (int i = 0; i < m; i++) ans[i] = -temp[i];
  reverse(w + 1, w + n + 1);
  for (int i = 0; i < m; i++) {
    int l = ql[i], r = qr[i];
    ql[i] = n + 1 - r;
    qr[i] = n + 1 - l;
  }
  for (int i = 1; i <= n; i++) val[i] = n + 1 - i;
  solve(temp);
  for (int i = 0; i < m; i++) ans[i] += temp[i];
  for (int i = 0; i < m; i++)
    printf("%lld%c", ans[i] + qr[i] - ql[i] + 1, " \n"[i + 1 == m]);
  return 0;
}