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 + 10;
int n, q;
int p[N];
struct que {
  int l, r, ids;
} qu[N];
int nums[N];
int L[N], R[N];
long long tree[N << 2], change[N << 2];
void opt1(int o, int l, int r, int pos, long long w) {
  if (l == r) {
    change[o] = (w == 0);
    tree[o] = w;
    return;
  }
  int mid = (l + r) >> 1;
  if (mid >= pos)
    opt1(o << 1, l, mid, pos, w);
  else
    opt1(o << 1 | 1, mid + 1, r, pos, w);
  tree[o] = tree[o << 1] + tree[o << 1 | 1];
  change[o] = change[o << 1] + change[o << 1 | 1];
}
long long opt2(int o, int l, int r, int L, int R, long long w) {
  if (L <= l && r <= R) return tree[o] + change[o] * w;
  int mid = (l + r) >> 1;
  long long re = 0;
  if (mid >= L) re += opt2(o << 1, l, mid, L, R, w);
  if (mid < R) re += opt2(o << 1 | 1, mid + 1, r, L, R, w);
  return re;
}
bool pd1(int i, int j) { return L[i] < L[j]; }
bool pd2(int i, int j) { return R[i] < R[j]; }
bool pd3(que i, que j) { return i.l < j.l; }
bool pd4(que i, que j) { return i.r < j.r; }
long long ans[N];
int st[N], top;
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &p[i]);
    nums[i] = i;
  }
  for (int i = 1; i <= n; ++i) {
    while (p[st[top]] < p[i] && top) --top;
    if (top)
      L[i] = st[top] + 1;
    else
      L[i] = 1;
    st[++top] = i;
  }
  top = 0;
  for (int i = n; i; --i) {
    while (p[st[top]] < p[i] && top) --top;
    if (top)
      R[i] = st[top] - 1;
    else
      R[i] = n;
    st[++top] = i;
  }
  for (int i = 1; i <= q; ++i) {
    qu[i].ids = i;
    scanf("%d", &qu[i].l);
  }
  for (int i = 1; i <= q; ++i) {
    scanf("%d", &qu[i].r);
    ans[i] += qu[i].r - qu[i].l + 1;
  }
  sort(nums + 1, nums + 1 + n, pd2);
  sort(qu + 1, qu + 1 + q, pd4);
  int now = n + 1;
  for (int i = 1; i <= n; ++i) opt1(1, 1, n, i, R[i]);
  for (int i = q; i; --i) {
    while (qu[i].r < R[nums[now - 1]] && now - 1 >= 1)
      opt1(1, 1, n, nums[now - 1], 0), --now;
    ans[qu[i].ids] += opt2(1, 1, n, qu[i].l, qu[i].r, qu[i].r);
  }
  sort(nums + 1, nums + 1 + n, pd1);
  sort(qu + 1, qu + 1 + q, pd3);
  now = 0;
  for (int i = 1; i <= n; ++i) opt1(1, 1, n, i, -(L[i]));
  for (int i = 1; i <= q; ++i) {
    while (qu[i].l > L[nums[now + 1]] && now + 1 <= n)
      opt1(1, 1, n, nums[now + 1], 0), ++now;
    ans[qu[i].ids] += opt2(1, 1, n, qu[i].l, qu[i].r, -qu[i].l);
  }
  for (int i = 1; i <= q; ++i) printf("%lld ", ans[i]);
}