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 MAXN = 1e6 + 7;
int a[MAXN], st[MAXN], d[MAXN][27], L[MAXN], R[MAXN];
long long fd[MAXN], f[MAXN], sf[MAXN], gd[MAXN], g[MAXN], sg[MAXN];
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  int top;
  st[top = 0] = 0;
  for (int i = 1; i <= n; i++) {
    while (top && a[st[top]] < a[i]) --top;
    fd[i] = top;
    f[i] = f[i - 1] + top + i - st[top];
    sf[i] = sf[i - 1] + f[i];
    st[++top] = i;
  }
  a[st[top = 0] = n + 1] = 0;
  for (int i = n; i >= 1; i--) {
    while (top && a[st[top]] < a[i]) --top;
    gd[i] = top;
    g[i] = g[i + 1] + st[top] - i + top;
    sg[i] = sg[i + 1] + g[i];
    st[++top] = i;
  }
  for (int i = 1; i <= n; i++) d[i][0] = i;
  for (int j = 1; (1 << j) <= n; j++)
    for (int i = 1; i + (1 << j) - 1 <= n; i++)
      if (a[d[i][j - 1]] > a[d[i + (1 << (j - 1))][j - 1]])
        d[i][j] = d[i][j - 1];
      else
        d[i][j] = d[i + (1 << (j - 1))][j - 1];
  for (int i = 1; i <= q; i++) scanf("%d", L + i);
  for (int i = 1; i <= q; i++) scanf("%d", R + i);
  int k;
  for (int i = 1; i <= q; i++) {
    if (L[i] > R[i]) swap(L[i], R[i]);
    k = 0;
    while (1 << (k + 1) <= R[i] - L[i] + 1) k++;
    if (a[d[L[i]][k]] > a[d[R[i] - (1 << k) + 1][k]])
      k = d[L[i]][k];
    else
      k = d[R[i] - (1 << k) + 1][k];
    long long ans = -fd[k] * (R[i] - k) - gd[k] * (k - L[i]) + 1;
    ans += f[R[i]] - f[k];
    ans += g[L[i]] - g[k];
    printf("%lld ", ans);
  }
  return 0;
}