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 + 10;
;
const int mod = 1e9 + 7;
int n, q;
int a[maxn], L[maxn], R[maxn];
long long lazy[maxn * 4], seg[maxn * 4];
void pushup(int l, int r, int v) {
  int mid = (l + r) >> 1;
  seg[v] = seg[(v << 1)] + (mid - l + 1) * lazy[(v << 1)];
  seg[v] += seg[(v << 1 | 1)] + (r - mid) * lazy[(v << 1 | 1)];
}
void pushdown(int v) {
  lazy[(v << 1)] += lazy[v];
  lazy[(v << 1 | 1)] += lazy[v];
  lazy[v] = 0;
}
void update(int ql, int qr, int l, int r, int v) {
  if (r < ql || l > qr) return;
  if (ql <= l && r <= qr) {
    lazy[v]++;
    return;
  }
  int mid = (l + r) >> 1;
  pushdown(v);
  update(ql, qr, l, mid, (v << 1));
  update(ql, qr, mid + 1, r, (v << 1 | 1));
  pushup(l, r, v);
}
long long query(int ql, int qr, int l, int r, int v) {
  if (r < ql || l > qr) return 0;
  if (ql <= l && r <= qr) return seg[v] + (r - l + 1) * lazy[v];
  int mid = (l + r) >> 1;
  pushdown(v);
  long long res = 0;
  res += query(ql, qr, l, mid, (v << 1));
  res += query(ql, qr, mid + 1, r, (v << 1 | 1));
  pushup(l, r, v);
  return res;
}
vector<int> l[maxn], r[maxn];
long long ans[maxn];
int x[maxn], y[maxn], stk[maxn];
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", &x[i]);
    l[x[i]].push_back(i);
  }
  for (int i = 1; i <= q; i++) {
    scanf("%d", &y[i]);
    r[y[i]].push_back(i);
  }
  a[n + 1] = a[0] = INT_MAX;
  int top = 0;
  stk[++top] = 0;
  for (int i = 1; i <= n; i++) {
    while (a[stk[top]] < a[i]) top--;
    L[i] = stk[top];
    stk[++top] = i;
  }
  top = 1;
  stk[top] = n + 1;
  for (int i = n; i >= 1; i--) {
    while (a[stk[top]] < a[i]) top--;
    R[i] = stk[top];
    stk[++top] = i;
  }
  for (int i = 1; i <= n; i++) {
    for (auto& u : l[i]) {
      ans[u] -= query(x[u], y[u], 1, n, 1);
    }
    update(L[i] + 1, R[i] - 1, 1, n, 1);
    for (auto& u : r[i]) {
      ans[u] += query(x[u], y[u], 1, n, 1);
    }
  }
  for (int i = 1; i <= q; i++) printf("%lld ", ans[i]);
  puts("");
  return 0;
}