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 N = 1e6 + 10;
int n, q, l[N], r[N], a[N];
long long ans[N];
vector<int> ls[N];
struct BIT {
  long long t[N];
  void clear() { memset(t, 0, sizeof t); }
  void add(int x, int val) {
    for (; x <= n + 1; x += x & -x) t[x] += val;
  }
  long long sum(int x) {
    long long res = 0;
    for (; x; x -= x & -x) res += t[x];
    return res;
  }
} tr[2];
void solve() {
  stack<int> s;
  s.push(n + 1);
  tr[0].clear();
  tr[1].clear();
  for (int i = (1); i <= (n); ++i) ls[i].clear();
  for (int i = (1); i <= (q); ++i) ls[l[i]].push_back(i);
  for (int i = n; i; --i) {
    while (a[s.top()] < a[i]) s.pop();
    int j = s.top();
    s.push(i);
    tr[0].add(i, -i);
    tr[0].add(j, j - 1);
    tr[1].add(i, 1);
    tr[1].add(j, -1);
    for (auto id : ls[i])
      ans[id] = ans[id] + tr[0].sum(r[id]) + r[id] * tr[1].sum(r[id]);
  }
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = (1); i <= (n); ++i) scanf("%d", a + i);
  a[0] = a[n + 1] = N;
  for (int i = (1); i <= (q); ++i) scanf("%d", l + i);
  for (int i = (1); i <= (q); ++i) scanf("%d", r + i);
  for (int i = (1); i <= (q); ++i) ans[i] = r[i] - l[i] + 1;
  solve();
  for (int i = (1); i <= (q); ++i)
    l[i] = n + 1 - l[i], r[i] = n + 1 - r[i], swap(l[i], r[i]);
  reverse(a, a + n + 2);
  solve();
  for (int i = (1); i <= (q); ++i) printf("%lld ", ans[i]);
}