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 qL[1000010], qR[1000010];
int L[1000010], R[1000010];
int p[1000010];
vector<pair<int, int> > vecL[1000010], vecR[1000010];
long long ans[1000010];
long long B[1000010];
void update(int u, int first) {
  while (u < 1000010) {
    B[u] += first;
    u += u & (-u);
  }
}
long long calc(int u) {
  long long ss = 0;
  while (u > 0) {
    ss += B[u];
    u -= u & (-u);
  }
  return ss;
}
long long tB[1000010];
void tupdate(int u, int first) {
  while (u < 1000010) {
    tB[u] += first;
    u += u & (-u);
  }
}
long long tcalc(int u) {
  long long ss = 0;
  while (u > 0) {
    ss += tB[u];
    u -= u & (-u);
  }
  return ss;
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
  for (int i = 1; i <= n; i++) {
    int st = i - 1;
    while (st && p[st] <= p[i]) {
      st = L[st];
    }
    L[i] = st;
  }
  R[n] = n + 1;
  for (int i = n - 1; i >= 1; i--) {
    int st = i + 1;
    while (st <= n && p[st] <= p[i]) {
      st = R[st];
    }
    R[i] = st;
  }
  for (int i = 1; i <= q; i++) {
    scanf("%d", &qL[i]);
  }
  for (int i = 1; i <= q; i++) {
    scanf("%d", &qR[i]);
  }
  for (int i = 1; i <= q; i++) {
    vecL[qL[i]].push_back(pair<int, int>(qR[i], i));
    vecR[qR[i]].push_back(pair<int, int>(qL[i], i));
  }
  for (int i = 1; i <= n; i++) {
    int first = L[i] + 1;
    tupdate(1, i - first);
    tupdate(first + 1, first - i);
    tupdate(first + 1, i);
    tupdate(i + 1, -i);
    update(first + 1, 1);
    update(i + 1, -1);
    for (auto V : vecR[i]) {
      int L = V.first, id = V.second;
      ans[id] += tcalc(L) - calc(L) * L;
    }
  }
  memset(B, 0, sizeof B);
  memset(tB, 0, sizeof tB);
  for (int i = n; i >= 1; i--) {
    int first = R[i] - 1;
    tupdate(first, first - i);
    tupdate(n + 1, i - first);
    tupdate(i, -i);
    tupdate(first, i);
    update(i, 1);
    update(first, -1);
    for (auto V : vecL[i]) {
      int R = V.first, id = V.second;
      ans[id] += tcalc(R) + calc(R) * R;
    }
  }
  for (int i = 1; i <= q; i++) {
    printf("%I64d ", ans[i] + qR[i] - qL[i] + 1);
  }
}