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 i64 = long long;
const int N = 1000000 + 7;
int n, q;
int a[N], l[N], r[N];
int sta[N], top;
std::vector<int> v[N];
i64 ans[N];
struct str {
  int l, r, i;
} e[N];
struct SegmentTree {
  i64 dat[N << 2];
  void init() { std::fill(dat + 1, dat + n * 4 + 1, 0); }
  void add(int p, int l, int r, int x, i64 y) {
    if (l < r) {
      int m = l + r >> 1;
      x <= m ? add(p << 1, l, m, x, y) : add(p << 1 | 1, m + 1, r, x, y);
    }
    dat[p] += y;
  }
  i64 sum(int p, int l, int r, int x, int y) {
    if (x <= l && r <= y) return dat[p];
    int m = l + r >> 1;
    i64 t = 0;
    if (x <= m) t += sum(p << 1, l, m, x, y);
    if (y > m) t += sum(p << 1 | 1, m + 1, r, x, y);
    return t;
  }
} trP, trR;
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", &e[i].l);
  for (int i = 1; i <= q; ++i) scanf("%d", &e[i].r);
  for (int i = 1; i <= q; ++i) ans[e[i].i = i] = e[i].r - e[i].l + 1;
  a[n + 1] = n + 1, sta[top = 1] = n + 1;
  for (int i = n; i; --i) {
    while (a[i] >= a[sta[top]]) --top;
    r[i] = sta[top] - 1, sta[++top] = i;
  }
  std::sort(e + 1, e + q + 1, [](str a, str b) { return a.r > b.r; });
  trP.init(), trR.init();
  for (int i = 1; i <= n; ++i) v[i].clear();
  for (int i = 1; i <= n; ++i) v[r[i]].push_back(i), trP.add(1, 1, n, i, r[i]);
  for (int i = n, j = 1; i; --i) {
    for (auto j : v[i]) trP.add(1, 1, n, j, -i), trR.add(1, 1, n, j, 1);
    for (; e[j].r == i && j <= q; ++j)
      ans[e[j].i] += trP.sum(1, 1, n, e[j].l, e[j].r) +
                     trR.sum(1, 1, n, e[j].l, e[j].r) * i;
  }
  a[0] = n + 1, sta[top = 1] = 0;
  for (int i = 1; i <= n; ++i) {
    while (a[i] >= a[sta[top]]) --top;
    l[i] = sta[top] + 1, sta[++top] = i;
  }
  std::sort(e + 1, e + q + 1, [](str a, str b) { return a.l < b.l; });
  trP.init(), trR.init();
  for (int i = 1; i <= n; ++i) v[i].clear();
  for (int i = 1; i <= n; ++i) v[l[i]].push_back(i), trP.add(1, 1, n, i, l[i]);
  for (int i = 1, j = 1; i <= n; ++i) {
    for (auto j : v[i]) trP.add(1, 1, n, j, -i), trR.add(1, 1, n, j, 1);
    for (; e[j].l == i && j <= q; ++j)
      ans[e[j].i] -= trP.sum(1, 1, n, e[j].l, e[j].r) +
                     trR.sum(1, 1, n, e[j].l, e[j].r) * i;
  }
  for (int i = 1; i <= q; ++i) printf("%lld ", ans[i]);
  return 0;
}