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 + 5;
long long sum[N * 4], f[N * 4];
void push_down(int l, int r, int root) {
  if (!f[root]) return;
  int mid = l + r >> 1;
  sum[root << 1] += (mid - l + 1) * f[root];
  sum[root << 1 | 1] += (r - mid) * f[root];
  f[root << 1] += f[root];
  f[root << 1 | 1] += f[root];
  f[root] = 0;
}
void update(int l, int r, int root, int ql, int qr, int v) {
  if (l >= ql && r <= qr) {
    sum[root] += (r - l + 1) * v;
    f[root]++;
    return;
  }
  push_down(l, r, root);
  int mid = l + r >> 1;
  if (mid >= ql) update(l, mid, root << 1, ql, qr, v);
  if (mid < qr) update(mid + 1, r, root << 1 | 1, ql, qr, v);
  sum[root] = sum[root << 1] + sum[root << 1 | 1];
}
long long query(int l, int r, int root, int ql, int qr) {
  if (l >= ql && r <= qr) return sum[root];
  push_down(l, r, root);
  int mid = l + r >> 1;
  long long ans = 0;
  if (mid >= ql) ans = query(l, mid, root << 1, ql, qr);
  if (mid < qr) ans += query(mid + 1, r, root << 1 | 1, ql, qr);
  return ans;
}
int p[N], l[N], r[N], lef[N], rig[N];
vector<int> ql[N], qr[N];
long long ans[N];
int st[N], top;
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  p[0] = p[n + 1] = 1e9;
  for (int i = 1; i <= m; i++) scanf("%d", &l[i]);
  for (int i = 1; i <= m; i++) scanf("%d", &r[i]);
  for (int i = 1; i <= m; i++) qr[l[i]].push_back(i), ql[r[i]].push_back(i);
  st[++top] = 0;
  for (int i = 1; i <= n; i++) {
    while (p[st[top]] < p[i]) top--;
    lef[i] = st[top];
    st[++top] = i;
  }
  top = 0;
  st[++top] = n + 1;
  for (int i = n; i; i--) {
    while (p[st[top]] < p[i]) top--;
    rig[i] = st[top];
    st[++top] = i;
  }
  for (int i = 1; i <= n; i++) {
    update(1, n, 1, lef[i] + 1, i, 1);
    for (auto j : ql[i]) ans[j] += query(1, n, 1, l[j], i);
  }
  memset(sum, 0, sizeof(sum));
  memset(f, 0, sizeof(f));
  for (int i = n; i; i--) {
    if (i + 1 <= rig[i] - 1) update(1, n, 1, i + 1, rig[i] - 1, 1);
    for (auto j : qr[i]) ans[j] += query(1, n, 1, i, r[j]);
  }
  for (int i = 1; i <= m; i++) printf("%lld%c", ans[i], " \n"[i == m]);
  return 0;
}