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 = 1000010;
long long ans[N], tag1[N << 2], tag2[N << 2];
int n, m, a[N], stk[N], nxt[N];
struct P {
  int l, r, id;
  long long ans;
} q[N];
bool operator<(const P &a, const P &b) { return a.l < b.l; }
void build(int x, int L, int R) {
  if (L == R) {
    tag1[x] = tag2[x] = 0;
    return;
  }
  int mid = (L + R) >> 1;
  build((x << 1), L, mid);
  build(((x << 1) | 1), mid + 1, R);
}
void push(int x) {
  if (tag1[x])
    tag1[(x << 1)] += tag1[x], tag1[((x << 1) | 1)] += tag1[x], tag1[x] = 0;
  if (tag2[x])
    tag2[(x << 1)] += tag2[x], tag2[((x << 1) | 1)] += tag2[x], tag2[x] = 0;
}
void mdf(int x, int L, int R, int l, int r, int k, int b) {
  if (L == l && r == R) {
    tag1[x] += k;
    tag2[x] += b;
    return;
  }
  int mid = (L + R) >> 1;
  push(x);
  if (r <= mid)
    mdf((x << 1), L, mid, l, r, k, b);
  else if (l > mid)
    mdf(((x << 1) | 1), mid + 1, R, l, r, k, b);
  else
    mdf((x << 1), L, mid, l, mid, k, b),
        mdf(((x << 1) | 1), mid + 1, R, mid + 1, r, k, b);
}
long long que(int x, int L, int R, int k) {
  if (L == R) return tag1[x] * k + tag2[x];
  int mid = (L + R) >> 1;
  push(x);
  if (k <= mid)
    return que((x << 1), L, mid, k);
  else
    return que(((x << 1) | 1), mid + 1, R, k);
}
void solve() {
  int top = 1;
  stk[1] = n + 1;
  a[n + 1] = n + 1;
  for (int i = n; i; i--) {
    while (top && a[stk[top]] < a[i]) top--;
    nxt[i] = stk[top];
    stk[++top] = i;
  }
  sort(q + 1, q + m + 1);
  int r = 0;
  build(1, 1, n);
  for (int i = (1); i <= (n); i++) {
    while (r < m && q[r + 1].l == i) r++, q[r].ans -= que(1, 1, n, q[r].r);
    mdf(1, 1, n, i, nxt[i] - 1, 1, -i);
    if (nxt[i] <= n) mdf(1, 1, n, nxt[i], n, 0, nxt[i] - i - 1);
  }
  for (int i = (1); i <= (m); i++) q[i].ans += que(1, 1, n, q[i].r);
}
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", &q[i].l), q[i].id = i;
  for (int i = (1); i <= (m); i++)
    scanf("%d", &q[i].r), q[i].ans = q[i].r - q[i].l + 1;
  solve();
  reverse(a + 1, a + n + 1);
  for (int i = (1); i <= (m); i++)
    q[i].l = n - q[i].l + 1, q[i].r = n - q[i].r + 1, swap(q[i].l, q[i].r);
  solve();
  for (int i = (1); i <= (m); i++) ans[q[i].id] = q[i].ans;
  for (int i = (1); i <= (m); i++) cout << ans[i] << ' ';
  return 0;
}