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 = 1000011;
int64_t op[N << 2];
int n, a[N], l[N], r[N], d[N], q, laz[N << 2];
int64_t dp[N];
pair<pair<int, int>, int> e[N];
int64_t ans[N];
bool ss(const pair<pair<int, int>, int> &u,
        const pair<pair<int, int>, int> &v) {
  return u.first.second < v.first.second;
}
void re(int id, int l, int r) {
  op[id] = laz[id] = 0;
  if (l == r) return;
  int mid = (l + r) >> 1;
  re(id << 1, l, mid), re(id << 1 | 1, mid + 1, r);
}
void update(int id, int l, int r, int u, int v) {
  if (l > v || r < u) return;
  if (u <= l && r <= v) {
    op[id] += r - l + 1;
    ++laz[id];
    return;
  }
  int mid = (l + r) >> 1, uu = id << 1, vv = uu | 1;
  if (laz[id]) {
    op[uu] += (mid - l + 1) * 1LL * laz[id], laz[uu] += laz[id];
    op[vv] += (r - mid) * 1LL * laz[id], laz[vv] += laz[id];
    laz[id] = 0;
  }
  update(uu, l, mid, u, v), update(vv, mid + 1, r, u, v);
  op[id] = op[uu] + op[vv];
}
int64_t get(int id, int l, int r, int u, int v) {
  if (l > v || r < u) return 0;
  if (u <= l && r <= v) return op[id];
  int mid = (l + r) >> 1, uu = id << 1, vv = uu | 1;
  if (laz[id]) {
    op[uu] += (mid - l + 1) * 1LL * laz[id], laz[uu] += laz[id];
    op[vv] += (r - mid) * 1LL * laz[id], laz[vv] += laz[id];
    laz[id] = 0;
  }
  int64_t tmp = get(uu, l, mid, u, v) + get(vv, mid + 1, r, u, v);
  op[id] = op[uu] + op[vv];
  return tmp;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  a[0] = a[n + 1] = 1000000000;
  for (int i = 1; i <= n; ++i)
    for (l[i] = i; a[l[i] - 1] < a[i]; l[i] = l[l[i] - 1])
      ;
  for (int i = n; i > 0; --i)
    for (r[i] = i; a[r[i] + 1] < a[i]; r[i] = r[r[i] + 1])
      ;
  for (int i = 1; i <= n; ++i) ++d[l[i]], --d[r[i] + 1];
  for (int i = 1; i <= n; ++i) d[i] += d[i - 1];
  for (int i = 1; i <= n; ++i) dp[i] = d[i] + dp[i - 1];
  for (int i = 1; i <= q; ++i) scanf("%d", &e[i].first.first);
  for (int i = 1; i <= q; ++i) scanf("%d", &e[i].first.second);
  for (int i = 1; i <= q; ++i) {
    e[i].second = i;
    ans[i] = dp[e[i].first.second] - dp[e[i].first.first - 1];
  }
  sort(e + 1, e + 1 + q);
  re(1, 1, n);
  for (int i = 1, j = 1; i <= q; ++i) {
    while (j < e[i].first.first) {
      update(1, 1, n, j, r[j]);
      ++j;
    }
    ans[e[i].second] -= get(1, 1, n, e[i].first.first, e[i].first.second);
  }
  sort(e + 1, e + 1 + q, ss);
  re(1, 1, n);
  for (int i = q, j = n; i > 0; --i) {
    while (j > e[i].first.second) {
      update(1, 1, n, l[j], j);
      --j;
    }
    ans[e[i].second] -= get(1, 1, n, e[i].first.first, e[i].first.second);
  }
  for (int i = 1; i <= q; ++i) printf("%I64d ", ans[i]);
}