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, m, ql[1000005], qr[1000005], a[1000005], L[1000005], stk[1000005], top;
long long ans[1000005];
vector<int> Q[1000005], G[1000005];
struct BIT {
  long long arr[1000005];
  void clear() { memset(arr, 0, sizeof arr); }
  void upd(int i, int d) {
    for (; i <= n; i += i & -i) arr[i] += d;
  }
  long long qsum(int i) {
    long long s = 0;
    for (; i; i -= i & -i) s += arr[i];
    return s;
  }
  long long ask(int l, int r) { return qsum(r) - qsum(l - 1); }
} T[2];
void solve() {
  T[0].clear(), T[1].clear(), top = 0;
  for (int i = 1; i <= n; i++) Q[i].clear(), G[i].clear();
  for (int i = 1; i <= m; i++)
    ans[i] += 1ll * (ql[i] + qr[i]) * (qr[i] - ql[i] + 1) / 2,
        Q[ql[i]].push_back(i);
  for (int i = 1; i <= n; i++) {
    while (top && a[stk[top]] < a[i]) top--;
    L[i] = stk[top] + 1, stk[++top] = i;
    T[0].upd(i, L[i]), G[L[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    for (int x : Q[i]) ans[x] -= T[0].ask(i, qr[x]) + T[1].ask(i, qr[x]) * i;
    for (int x : G[i]) T[0].upd(x, -L[x]), T[1].upd(x, 1);
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= m; i++) scanf("%d", &ql[i]);
  for (int i = 1; i <= m; i++) scanf("%d", &qr[i]);
  solve();
  reverse(a + 1, a + 1 + n);
  for (int i = 1; i <= m; i++)
    ql[i] = n - ql[i] + 1, qr[i] = n - qr[i] + 1, swap(ql[i], qr[i]);
  solve();
  for (int i = 1; i <= m; i++)
    printf("%I64d%c", ans[i] + qr[i] - ql[i] + 1, i == m ? 10 : 32);
}