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, idx;
} q[1000005];
bool cmp(const node &a, const node &b) {
  return a.r == b.r ? a.l < b.l : a.r < b.r;
}
int n, Q;
long long s[1000005 << 2], tag[1000005 << 2], ans[1000005], t[1000005];
int sta[1000005], top, a[1000005];
void Update(int L, int R, int v, int l, int r, int rt) {
  if (L > R) return;
  s[rt] += (long long)(min(R, r) - max(l, L) + 1) * v;
  if (L <= l && r <= R) {
    tag[rt] += v;
    return;
  }
  int m = (l + r) >> 1;
  if (L <= m) Update(L, R, v, l, m, rt << 1);
  if (m < R) Update(L, R, v, m + 1, r, rt << 1 | 1);
}
long long query(int L, int R, int l, int r, int rt) {
  if (L <= l && r <= R) return s[rt];
  int m = (l + r) >> 1;
  long long ret = tag[rt] * (min(r, R) - max(l, L) + 1);
  if (L <= m) ret += query(L, R, l, m, rt << 1);
  if (m < R) ret += query(L, R, m + 1, r, rt << 1 | 1);
  return ret;
}
void print(int now, int l, int r, int rt) {
  if (l == r) {
    printf("%d ", s[rt] + now);
    return;
  }
  int m = (l + r) >> 1;
  print(now + tag[rt], l, m, rt << 1);
  print(now + tag[rt], m + 1, r, rt << 1 | 1);
}
int main() {
  scanf("%d%d", &n, &Q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= Q; i++) q[i].idx = i, scanf("%d", &q[i].l);
  for (int i = 1; i <= Q; i++) scanf("%d", &q[i].r);
  sort(q + 1, q + Q + 1, cmp);
  for (int i = 1, j = 1; i <= n; i++) {
    while (top && a[sta[top]] < a[i]) {
      int p = sta[top - 1];
      Update(p + 1, sta[top] - 1, 1, 1, n, 1);
      Update(sta[top], sta[top], i - sta[top], 1, n, 1);
      top--;
    }
    sta[++top] = i, t[top] = t[top - 1] + sta[top - 1];
    while (i == q[j].r) {
      int p = lower_bound(sta + 1, sta + top + 1, q[j].l) - sta - 1;
      ans[q[j].idx] = (long long)(top - p) * i -
                      (t[top] - t[p + 1] + q[j].l - 1) +
                      query(q[j].l, i, 1, n, 1),
      j++;
    }
  }
  for (int i = 1; i <= Q; i++) printf("%lld ", ans[i]);
  puts("");
}