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 e {
  long long p, id;
} eve[3000005];
long long n, q, i, po, fd[1000005], tmp, qu[1000005], hd, tl, a[1000005],
    l[1000005], r[1000005], ans[1000005], res2[1000005], cnt = 0,
                                                         tr[2][1000005];
const long long inf = 1e18;
long long lowbit(long long p) { return p & -p; }
void add(long long t, long long pos, long long v) {
  while (pos <= n) {
    tr[t][pos] += v;
    pos += lowbit(pos);
  }
}
long long ask(long long t, long long pos) {
  long long ans = 0;
  while (pos) {
    ans += tr[t][pos];
    pos -= lowbit(pos);
  }
  return ans;
}
bool cmp(e a, e b) { return abs(a.p) < abs(b.p); }
void solve() {
  long long i, u, v;
  a[0] = inf;
  hd = tl = 1;
  for (i = 0; i <= n; i++) {
    while (hd < tl && a[qu[tl - 1]] < a[i]) tl--;
    qu[tl++] = i;
    if (i) fd[i] = qu[tl - 2] + 1;
  }
  cnt = 0;
  po = 1;
  for (i = 1; i <= q; i++) {
    eve[++cnt] = (e){-(l[i] - 1), i};
    eve[++cnt] = (e){r[i], i};
  }
  sort(eve + 1, eve + cnt + 1, cmp);
  for (i = 1; i <= cnt; i++) {
    while (po <= n && po <= abs(eve[i].p)) {
      add(0, fd[po], 1);
      add(1, fd[po], fd[po]);
      po++;
    }
    if (eve[i].p < 0) {
      res2[eve[i].id] = ask(1, -eve[i].p);
      ans[eve[i].id] -=
          (1LL - eve[i].p) * (-eve[i].p) / 2LL - res2[eve[i].id] - eve[i].p;
    } else {
      v = ask(1, eve[i].p);
      ans[eve[i].id] += (1LL + eve[i].p) * eve[i].p / 2LL - v + eve[i].p;
      u = ask(0, l[eve[i].id] - 1);
      v = ask(1, l[eve[i].id] - 1);
      ans[eve[i].id] -=
          (u - l[eve[i].id] + 1) * l[eve[i].id] - v + res2[eve[i].id];
    }
  }
}
int main() {
  scanf("%lld%lld", &n, &q);
  for (i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (i = 1; i <= q; i++) scanf("%lld", &l[i]);
  for (i = 1; i <= q; i++) scanf("%lld", &r[i]);
  solve();
  memset(tr, 0, sizeof(tr));
  memset(res2, 0, sizeof(res2));
  for (i = 1; i <= n / 2; i++) swap(a[i], a[n - i + 1]);
  for (i = 1; i <= q; i++) tmp = l[i], l[i] = n - r[i] + 1, r[i] = n - tmp + 1;
  solve();
  for (i = 1; i <= q; i++) printf("%lld ", ans[i] - r[i] + l[i] - 1);
  return 0;
}