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;
inline int read() {
  int x = 0, f = 1;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f = -1;
  } while (ch < '0' || ch > '9');
  do {
    x = x * 10 + ch - '0';
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  return x * f;
}
inline pair<long long, int> operator+(const pair<long long, int>& t1,
                                      const pair<long long, int>& t2) {
  return make_pair(t1.first + t2.first, t1.second + t2.second);
}
inline pair<long long, int> operator-(const pair<long long, int>& t1,
                                      const pair<long long, int>& t2) {
  return make_pair(t1.first - t2.first, t1.second - t2.second);
}
const int N = 1e6 + 10;
struct node {
  int l, r;
} q[N];
int n, nq, a[N], lb[N], rb[N], s[N], ls;
long long tans[N];
vector<int> lnum[N], rnum[N], ql[N], qr[N];
pair<long long, int> T[N];
void qadd(int x, pair<long long, int> v) {
  while (x <= n) T[x] = T[x] + v, x += ((x) & -(x));
}
pair<long long, int> qsum(int x) {
  pair<long long, int> ans = make_pair(0, 0);
  while (x) ans = ans + T[x], x -= ((x) & -(x));
  return ans;
}
int main() {
  scanf("%d%d", &n, &nq);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) {
    while (ls && a[s[ls]] < a[i]) rb[s[ls--]] = i - 1;
    lb[i] = s[ls] + 1;
    s[++ls] = i;
  }
  while (ls) rb[s[ls--]] = n;
  for (int i = 1; i <= nq; ++i) scanf("%d", &q[i].l);
  for (int i = 1; i <= nq; ++i) scanf("%d", &q[i].r);
  for (int i = 1; i <= nq; ++i) {
    ql[q[i].l].push_back(i);
    qr[q[i].r].push_back(i);
  }
  for (int i = 1; i <= n; ++i) {
    lnum[lb[i]].push_back(i);
    qadd(i, make_pair(lb[i], 0));
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < lnum[i].size(); ++j) qadd(lnum[i][j], make_pair(-i, 1));
    for (int j = 0; j < ql[i].size(); ++j) {
      int id = ql[i][j], l = q[id].l, r = q[id].r;
      pair<long long, int> tmp = qsum(r) - qsum(l - 1);
      tans[id] -= tmp.first + 1ll * i * tmp.second;
    }
  }
  memset(T, 0, sizeof(T));
  for (int i = 1; i <= n; ++i) {
    rnum[rb[i]].push_back(i);
    qadd(i, make_pair(rb[i], 0));
  }
  for (int i = n; i >= 1; --i) {
    for (int j = 0; j < rnum[i].size(); ++j) qadd(rnum[i][j], make_pair(-i, 1));
    for (int j = 0; j < qr[i].size(); ++j) {
      int id = qr[i][j], l = q[id].l, r = q[id].r;
      pair<long long, int> tmp = qsum(r) - qsum(l - 1);
      tans[id] += tmp.first + 1ll * i * tmp.second;
    }
  }
  for (int i = 1; i <= nq; ++i)
    printf("%lld%c", tans[i] + q[i].r - q[i].l + 1ll, i == nq ? '\n' : ' ');
  return 0;
}