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 + 5;
int n, m, a[N], l[N], r[N], L[N], R[N];
int64_t it[N * 4], lz[N * 4], ans[N];
vector<int> ql[N], qr[N];
void Down(int k, int l, int r, int m) {
  lz[((k) << 1)] += lz[k];
  it[((k) << 1)] += lz[k] * (m - l + 1);
  lz[((k) << 1) + 1] += lz[k];
  it[((k) << 1) + 1] += lz[k] * (r - m);
  lz[k] = 0;
}
void update(int k, int l, int r, int u, int v) {
  if (v < l || r < u) return;
  if (u <= l && r <= v) {
    lz[k] += 1;
    it[k] += r - l + 1;
  } else {
    int m = (l + r) >> 1;
    Down(k, l, r, m);
    update(((k) << 1), l, m, u, v);
    update(((k) << 1) + 1, m + 1, r, u, v);
    it[k] = it[((k) << 1)] + it[((k) << 1) + 1];
  }
}
int64_t get(int k, int l, int r, int u, int v) {
  if (v < l || r < u) return 0;
  if (u <= l && r <= v) return it[k];
  int m = (l + r) >> 1;
  Down(k, l, r, m);
  return get(((k) << 1), l, m, u, v) + get(((k) << 1) + 1, m + 1, r, u, v);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (int i = 1; i <= m; ++i) scanf("%d", l + i);
  for (int i = 1; i <= m; ++i) scanf("%d", r + i);
  for (int i = 1; i <= n; ++i)
    for (L[i] = i - 1; L[i] >= 1 && a[L[i]] < a[i]; L[i] = L[L[i]])
      ;
  for (int i = n; i >= 1; --i)
    for (R[i] = i + 1; R[i] <= n && a[R[i]] < a[i]; R[i] = R[R[i]])
      ;
  for (int i = 1; i <= m; ++i) {
    ql[l[i]].push_back(i);
    qr[r[i]].push_back(i);
    ans[i] = r[i] - l[i] + 1;
  }
  memset(it, 0, sizeof(it));
  memset(lz, 0, sizeof(lz));
  for (int i = n; i >= 1; --i) {
    if (i + 1 < R[i]) update(1, 1, n, i + 1, R[i] - 1);
    for (int j : ql[i]) ans[j] += get(1, 1, n, i + 1, r[j]);
  }
  memset(it, 0, sizeof(it));
  memset(lz, 0, sizeof(lz));
  for (int i = 1; i <= n; ++i) {
    if (i - 1 > L[i]) update(1, 1, n, L[i] + 1, i - 1);
    for (int j : qr[i]) ans[j] += get(1, 1, n, l[j], i - 1);
  }
  for (int i = 1; i <= m; ++i) printf("%I64d ", ans[i]);
}