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 + 10;
int stk[N], top;
inline int lowbit(int i) { return i & -i; }
int n, q, a[N];
struct BIT {
  long long dat[N];
  long long sum(int i) {
    long long res = 0;
    for (; i > 0; i -= lowbit(i)) res += dat[i];
    return res;
  }
  void add(int i, long long x) {
    for (; i <= n; i += lowbit(i)) dat[i] += x;
  }
} T0, T1;
long long sum(int i) { return T0.sum(i) + T1.sum(i) * i; }
void Add(int l, int r) {
  T0.add(l, r + 1);
  T1.add(l, -1);
  T1.add(r + 1, 1);
  T0.add(r + 1, -(r + 1));
}
long long acc[N];
vector<pair<int, int> > qry[N];
int L[N], R[N];
long long ans[N];
int Find(int x) {
  int l = 0, r = top;
  while (l + 1 < r) {
    int mid = (l + r) / 2;
    if (stk[mid] >= x)
      r = mid;
    else
      l = mid;
  }
  return r;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", a + 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++) {
    qry[R[i]].push_back(pair<int, int>(L[i], i));
  }
  for (int i = 1; i <= n; i++) {
    while (top > 0 && a[stk[top]] < a[i]) {
      long long val =
          acc[0] - acc[stk[top]] - 1LL * (top - 1) * (i - 1 - stk[top] + 1);
      T0.add(stk[top - 1] + 1, val);
      T0.add(stk[top] + 1, -val);
      --top;
    }
    Add(stk[top] + 1, i - 1);
    T0.add(1, i - stk[top] - 1);
    T0.add(stk[top] + 1, -i + stk[top] + 1);
    stk[++top] = i;
    acc[i] = acc[0];
    acc[0] += top;
    for (auto p : qry[i]) {
      int l = p.first;
      long long x = Find(l);
      ans[p.second] =
          sum(l) + acc[0] - acc[stk[x]] - (x - 1) * (i - stk[x] + 1);
    }
  }
  for (int i = 1; i <= q; i++) printf("%lld%c", ans[i], " \n"[i == q]);
}