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 + 7;
long long bit[2][N], ans[N];
int n, m, i, j, a[N], l[N], r[N], q[N], top;
vector<int> e[N];
void add(int p, int x, int v) {
  while (x <= n + 3) bit[p][x] += v, x += x & -x;
}
long long sum(int p, int x, long long res = 0) {
  while (x) res += bit[p][x], x -= x & -x;
  return res;
}
void go() {
  for (i = 1; i <= n; ++i) e[i].clear();
  memset(bit, 0, sizeof(bit));
  top = 0;
  for (i = 1; i <= m; ++i) e[l[i]].push_back(i);
  for (i = n; i >= 1; --i) {
    while (top && a[q[top]] < a[i]) top--;
    j = n + 1;
    if (top) j = q[top];
    q[++top] = i;
    add(0, i, -(i - 1));
    add(0, j, i - 1);
    add(1, i, 1);
    add(1, j, -1);
    add(0, j, j - i);
    for (auto& id : e[i])
      ans[id] += sum(0, r[id]), ans[id] += r[id] * sum(1, r[id]);
  }
}
int main() {
  for (scanf("%d%d", &n, &m), i = 1; i <= n; ++i) scanf("%d", a + i);
  for (i = 1; i <= m; ++i) scanf("%d", l + i);
  for (i = 1; i <= m; ++i) scanf("%d", r + i);
  go();
  for (i = 1; i <= m; ++i)
    l[i] = n - l[i] + 1, r[i] = n - r[i] + 1, swap(l[i], r[i]);
  reverse(a + 1, a + n + 1);
  go();
  for (i = 1; i <= m; ++i)
    printf("%lld%c", ans[i] - (r[i] - l[i] + 1), i == m ? '\n' : ' ');
}