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;
long long a[1000010];
long long l[1000010], r[1000010];
vector<pair<int, int>> g1[1000010], g2[1000010];
int n, q;
long long ans[1000010];
const int N = 1000000;
long long tree[N * 4];
long long add[N * 4];
int cnt_v;
void build(int n) {
  cnt_v = 1;
  while (cnt_v <= n) {
    cnt_v <<= 1;
  }
  memset(tree, 0, sizeof(tree));
  memset(add, 0, sizeof(add));
}
void push(int x, int lx, int rx) {
  int len = (rx - lx) / 2;
  int mid = (lx + rx) / 2;
  tree[x * 2 + 1] += add[x] * (mid - lx + 1);
  add[x * 2 + 1] += add[x];
  tree[x * 2 + 2] += add[x] * (rx - mid);
  add[x * 2 + 2] += add[x];
  add[x] = 0;
}
void upd(int x, int lx, int rx, int l, int r) {
  if (lx > rx) return;
  if (l > rx || r < lx) return;
  if (lx >= l && rx <= r) {
    tree[x] += (rx - lx + 1);
    add[x]++;
    return;
  }
  {
    push(x, lx, rx);
    upd(x * 2 + 1, lx, (lx + rx) / 2, l, r);
    upd(x * 2 + 2, (lx + rx) / 2 + 1, rx, l, r);
    tree[x] = tree[x * 2 + 1] + tree[x * 2 + 2];
  }
}
long long get(int x, int lx, int rx, int l, int r) {
  if (lx > rx) return 0;
  if (l > rx || r < lx) return 0;
  if (lx >= l && rx <= r) {
    return tree[x];
  } else {
    push(x, lx, rx);
    return get(x * 2 + 1, lx, (lx + rx) / 2, l, r) +
           get(x * 2 + 2, (lx + rx) / 2 + 1, rx, l, r);
  }
}
void init() { build(n); }
void upd(long long l, long long r, long long val) { upd(0, 1, cnt_v, l, r); }
long long get_sum(long long l, long long r) {
  long long t = get(0, 0, cnt_v, l, r);
  return t;
}
pair<int, int> st[1000040];
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i < n + 1; ++i) {
    scanf("%lld", &a[i]);
  }
  for (int i = 0; i < q; ++i) {
    scanf("%lld", &l[i]);
  }
  for (int i = 0; i < q; ++i) {
    scanf("%lld", &r[i]);
  }
  for (int i = 0; i < q; ++i) {
    g1[l[i]].push_back(make_pair(r[i], i));
    g2[r[i]].push_back(make_pair(l[i], i));
  }
  init();
  stack<pair<long long, int>> st;
  st.push(make_pair(10000000, 0));
  for (int i = 1; i < n + 1; ++i) {
    while (!st.empty() && a[i] > st.top().first) {
      st.pop();
    }
    long long s = st.top().second + 1;
    long long e = i;
    upd(s, e, 1);
    st.push(make_pair(a[i], i));
    for (int j = 0; j < g2[i].size(); ++j) {
      long long s = g2[i][j].first;
      int id = g2[i][j].second;
      ans[id] += get_sum(s, i);
    }
  }
  init();
  while (!st.empty()) st.pop();
  st.push(make_pair(10000000, n + 1));
  for (int i = n; i > 0; --i) {
    while (!st.empty() && a[i] > st.top().first) {
      st.pop();
    }
    long long s = i;
    long long e = st.top().second - 1;
    upd(s, e, 1);
    st.push(make_pair(a[i], i));
    for (int j = 0; j < g1[i].size(); ++j) {
      long long e = g1[i][j].first;
      int id = g1[i][j].second;
      ans[id] += get_sum(i, e);
    }
  }
  for (int i = 0; i < q; ++i) {
    printf("%lld ", ans[i] - (r[i] - l[i] + 1));
  }
  cout << endl;
}