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 = 1000000 + 10;
const int M = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
const int oo = 1000000000;
int n, m, v[N], l, r, lz[4 * N];
pair<pair<int, int>, int> q[N];
long long seg[4 * N], ans[N];
void fix(int u, int s, int e) {
  seg[u] += 1ll * lz[u] * (e - s + 1);
  if (s - e) {
    lz[u * 2] += lz[u];
    lz[u * 2 + 1] += lz[u];
  }
  lz[u] = 0;
}
void update(int u, int s, int e) {
  fix(u, s, e);
  if (s > r || e < l) return;
  if (s >= l && e <= r) {
    ++lz[u];
    fix(u, s, e);
    return;
  }
  update(u * 2, s, (s + e) / 2);
  update(u * 2 + 1, (s + e) / 2 + 1, e);
  seg[u] = seg[u * 2] + seg[u * 2 + 1];
}
long long get(int u, int s, int e) {
  fix(u, s, e);
  if (s > r || e < l) return 0;
  if (s >= l && e <= r) return seg[u];
  return get(u * 2, s, (s + e) / 2) + get(u * 2 + 1, (s + e) / 2 + 1, e);
}
void go() {
  stack<int> st;
  for (int i = 1, j = 0; i <= n; ++i) {
    l = 1;
    r = i;
    while (!st.empty() && v[i] > v[st.top()]) st.pop();
    if (!st.empty()) l = st.top() + 1;
    st.push(i);
    update(1, 1, n);
    while (j < m && q[j].first.first == i) {
      l = q[j].first.second;
      r = q[j].first.first;
      ans[q[j++].second] += get(1, 1, n);
    }
  }
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) scanf("%d", &v[i]);
  for (int i = 0; i < m; ++i) {
    q[i].second = i;
    scanf("%d", &q[i].first.second);
  }
  for (int i = 0; i < m; ++i) {
    scanf("%d", &q[i].first.first);
    ans[i] = -(q[i].first.first - q[i].first.second + 1);
  }
  sort(q, q + m);
  go();
  for (int i = 0; i < m; ++i) {
    swap(q[i].first.first, q[i].first.second);
    q[i].first.first = n + 1 - q[i].first.first;
    q[i].first.second = n + 1 - q[i].first.second;
  }
  sort(q, q + m);
  reverse(v + 1, v + n + 1);
  memset(seg, 0, sizeof(seg));
  memset(lz, 0, sizeof(lz));
  go();
  for (int i = 0; i < m; ++i) printf("%lld ", ans[i]);
  return 0;
}