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;
int n, q, a[1000005], maxp[26][1000005], lg2[1000005], st[1000005], top;
int pre[1000005], nxt[1000005], l[1000005], r[1000005];
long long dis1[1000005], dis2[1000005], f1[1000005], f2[1000005], s1[1000005],
    s2[1000005];
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    maxp[0][i] = i;
  }
  for (int i = 2; i <= n; i++) lg2[i] = lg2[i >> 1] + 1;
  for (int i = 1; i <= lg2[n]; i++)
    for (int j = 1; j <= n; j++)
      maxp[i][j] = a[maxp[i - 1][j]] > a[maxp[i - 1][j + (1 << (i - 1))]]
                       ? maxp[i - 1][j]
                       : maxp[i - 1][j + (1 << (i - 1))];
  for (int i = 1; i <= n; i++) {
    while (top && a[st[top]] < a[i]) top--;
    pre[i] = st[top];
    st[++top] = i;
    f1[i] = f1[pre[i]] + 1;
    dis1[i] = i - pre[i];
    s1[i] = s1[i - 1] + f1[i] + dis1[i];
  }
  top = 0;
  st[0] = n + 1;
  for (int i = n; i >= 1; i--) {
    while (top && a[st[top]] < a[i]) top--;
    nxt[i] = st[top];
    st[++top] = i;
    f2[i] = f2[nxt[i]] + 1;
    dis2[i] = nxt[i] - i;
    s2[i] = s2[i + 1] + f2[i] + dis2[i];
  }
  for (int i = 1; i <= q; i++) scanf("%d", &l[i]);
  for (int i = 1; i <= q; i++) scanf("%d", &r[i]);
  for (int i = 1; i <= q; i++) {
    int d = lg2[r[i] - l[i] + 1];
    int p = a[maxp[d][l[i]]] > a[maxp[d][r[i] - (1 << d) + 1]]
                ? maxp[d][l[i]]
                : maxp[d][r[i] - (1 << d) + 1];
    printf("%lld ", 1 + s1[r[i]] - s1[p] - f1[p] * (r[i] - p) + s2[l[i]] -
                        s2[p] - f2[p] * (p - l[i]));
  }
  printf("\n");
  return 0;
}