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 + 6;
long long t[maxn << 2], lz[maxn << 2];
int a[maxn], n, m;
int l[maxn], r[maxn];
vector<int> L[maxn], R[maxn];
long long ans[maxn];
int nge[maxn], pge[maxn];
void push(int u, int b, int e) {
  t[u] += lz[u] * (e - b + 1);
  if (b != e) {
    lz[u << 1] += lz[u];
    lz[u << 1 | 1] += lz[u];
  }
  lz[u] = 0;
}
void update(int u, int b, int e, int l, int r, long long val) {
  if (b > r or e < l) return;
  if (l <= b and e <= r) {
    lz[u] += val;
    push(u, b, e);
    return;
  }
  if (lz[u]) push(u, b, e);
  int mid = b + e >> 1;
  update(u << 1, b, mid, l, r, val);
  update(u << 1 | 1, mid + 1, e, l, r, val);
  t[u] = t[u << 1] + t[u << 1 | 1];
}
long long query(int u, int b, int e, int l, int r) {
  if (b > r or e < l) return 0;
  if (lz[u]) push(u, b, e);
  if (l <= b and e <= r) return t[u];
  int mid = b + e >> 1;
  return query(u << 1, b, mid, l, r) + query(u << 1 | 1, mid + 1, e, l, r);
}
int main(int argc, char const *argv[]) {
  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", l + i);
    L[l[i]].emplace_back(i);
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d", r + i);
    R[r[i]].emplace_back(i);
  }
  a[0] = a[n + 1] = n + 1;
  stack<int> stk;
  stk.push(0);
  for (int i = 1; i <= n; i++) {
    while (stk.size() and a[stk.top()] <= a[i]) stk.pop();
    pge[i] = stk.top();
    stk.push(i);
  }
  while (stk.size()) stk.pop();
  stk.push(n + 1);
  for (int i = n; i >= 1; i--) {
    while (stk.size() and a[stk.top()] <= a[i]) stk.pop();
    nge[i] = stk.top();
    stk.push(i);
  }
  for (int i = 1; i <= n; i++) {
    update(1, 1, n, pge[i] + 1, i, +1);
    for (int j : R[i]) {
      ans[j] += query(1, 1, n, l[j], r[j]);
    }
  }
  memset(t, 0, sizeof t);
  memset(lz, 0, sizeof lz);
  for (int i = n; i >= 1; i--) {
    update(1, 1, n, i, nge[i] - 1, +1);
    for (int j : L[i]) {
      ans[j] += query(1, 1, n, l[j], r[j]);
    }
  }
  for (int i = 1; i <= m; i++) {
    ans[i] -= (r[i] - l[i] + 1);
  }
  for (int i = 1; i <= m; i++) {
    printf("%lld ", ans[i]);
  }
  puts("");
  return 0;
}