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;
struct node {
  int l, r;
  long long sum, tag;
} tree[1000010 << 2];
int n, q, len, top;
int a[1000010], lind[1000010], rind[1000010], s[1000010], l[1000010],
    r[1000010];
vector<pair<int, int> > lq[1000010], rq[1000010];
long long ans[1000010];
inline void prework() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  a[0] = a[n + 1] = 2 * 1e9;
  top = 0;
  s[++top] = 0;
  for (int i = 1; i <= n; i++) {
    while (a[i] > a[s[top]] && top > 0) top--;
    s[++top] = i;
    lind[i] = s[top - 1];
  }
  top = 0;
  s[++top] = n + 1;
  for (int i = n; i >= 1; i--) {
    while (a[i] > a[s[top]] && top > 0) top--;
    s[++top] = i;
    rind[i] = s[top - 1];
  }
  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++) {
    lq[r[i]].push_back(make_pair(l[i], i));
    rq[l[i]].push_back(make_pair(r[i], i));
  }
}
inline void build(int k, int l, int r) {
  tree[k].l = l;
  tree[k].r = r;
  tree[k].sum = 0;
  tree[k].tag = 0;
  if (l == r) return;
  int mid = (l + r) >> 1;
  build(k << 1, l, mid);
  build(k << 1 | 1, mid + 1, r);
}
inline long long query(int k, int l, int r) {
  if (tree[k].l == l && tree[k].r == r) return tree[k].sum;
  int mid = (tree[k].l + tree[k].r) >> 1;
  long long res = (r - l + 1) * tree[k].tag;
  if (r <= mid)
    res += query(k << 1, l, r);
  else if (l > mid)
    res += query(k << 1 | 1, l, r);
  else
    res += query(k << 1, l, mid) + query(k << 1 | 1, mid + 1, r);
  return res;
}
inline void add(int k, int l, int r, int val) {
  tree[k].sum += (r - l + 1) * val;
  if (tree[k].l == l && tree[k].r == r) {
    tree[k].tag += val;
    return;
  }
  int mid = (tree[k].l + tree[k].r) >> 1;
  if (r <= mid)
    add(k << 1, l, r, val);
  else if (l > mid)
    add(k << 1 | 1, l, r, val);
  else
    add(k << 1, l, mid, val), add(k << 1 | 1, mid + 1, r, val);
}
inline void mainwork() {
  int len;
  build(1, 1, n);
  for (int i = 1; i <= n; i++) {
    add(1, lind[i] + 1, i, 1);
    len = lq[i].size();
    for (int j = 0; j < len; j++)
      ans[lq[i][j].second] += query(1, lq[i][j].first, i);
  }
  build(1, 1, n);
  for (int i = n; i >= 1; i--) {
    if (i + 1 <= rind[i] - 1) add(1, i + 1, rind[i] - 1, 1);
    len = rq[i].size();
    for (int j = 0; j < len; j++)
      ans[rq[i][j].second] += query(1, i, rq[i][j].first);
  }
}
inline void print() {
  for (int i = 1; i <= q; i++) printf("%lld ", ans[i]);
}
int main() {
  prework();
  mainwork();
  print();
  return 0;
}