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;
inline int read() {
  char c = getchar();
  int x = 0;
  while (c < '0' || c > '9') {
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + (c & 15);
    c = getchar();
  }
  return x;
}
struct Node {
  int val;
  int pos;
  Node(int v, int p) {
    val = v;
    pos = p;
  }
  Node() {}
};
struct Query {
  int id;
  int l;
  int r;
};
int cmp(Query q1, Query q2) { return q1.l < q2.l; }
Node a[maxn];
Query q[maxn];
long long bit1[maxn];
long long bit2[maxn];
long long ans[maxn];
int p[maxn];
vector<int> v[maxn];
int tail = 0;
int n, m;
int add(Node node) {
  while (tail && a[tail - 1].val < node.val) {
    tail--;
  }
  a[tail++] = node;
  if (tail == 1) {
    return node.pos;
  } else {
    return a[tail - 2].pos;
  }
}
int lowbit(int x) { return x & (-x); }
void add1(int x, int i) {
  while (i <= n) {
    bit1[i] += x;
    i += lowbit(i);
  }
}
void add2(int x, int i) {
  while (i <= n) {
    bit2[i] += x;
    i += lowbit(i);
  }
}
long long sum1(int i) {
  long long s = 0;
  while (i > 0) {
    s += bit1[i];
    i -= lowbit(i);
  }
  return s;
}
long long sum2(int i) {
  long long s = 0;
  while (i > 0) {
    s += bit2[i];
    i -= lowbit(i);
  }
  return s;
}
void solve() {
  sort(q + 1, q + m + 1, cmp);
  tail = 0;
  for (int i = 1; i <= n; i++) {
    v[i].clear();
    bit1[i] = 0;
    bit2[i] = 0;
  }
  for (int i = 1; i <= n; i++) {
    int pos = add(Node(p[i], i));
    if (pos == i) {
      add1(pos, i);
    } else {
      add1(i - pos, i);
      add2(1, i);
      v[pos].push_back(i);
    }
  }
  int cur = 0;
  for (int i = 1; i <= m; i++) {
    if (q[i].l > cur) {
      for (; cur < q[i].l - 1; cur++) {
        for (int k = 0; k < v[cur].size(); k++) {
          int l = v[cur][k];
          add1(cur, l);
          add2(-1, l);
        }
      }
    }
    ans[q[i].id] -=
        ((q[i].r - q[i].l + 1) - (sum2(q[i].r) - sum2(q[i].l - 1))) * cur;
    ans[q[i].id] += sum1(q[i].r) - sum1(q[i].l - 1);
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    p[i] = read();
  }
  for (int i = 1; i <= m; i++) {
    q[i].l = read();
  }
  for (int i = 1; i <= m; i++) {
    q[i].id = i;
    q[i].r = read();
    ans[i] -= (q[i].r - q[i].l + 1);
  }
  solve();
  for (int i = 1; 2 * i <= n; i++) {
    swap(p[i], p[n - i + 1]);
  }
  for (int i = 1; i <= m; i++) {
    int t = q[i].r;
    q[i].r = n - q[i].l + 1;
    q[i].l = n - t + 1;
  }
  solve();
  for (int i = 1; i <= m; i++) printf("%lld\n", ans[i]);
  return 0;
}