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 = 1000007;
const long long mod = 1000000007;
const long long ool = 1e18 + 7;
int n;
class Bits {
 public:
  inline int lowbit(int x) { return x & (-x); }
  long long c[maxn];
  void modify(int x, int v) {
    for (int i = x; i && i <= n; i += lowbit(i)) c[i] += v;
  }
  inline long long sum(int x) {
    long long s = 0;
    for (int i = x; i; i -= lowbit(i)) s += c[i];
    return s;
  }
  inline long long qury(int l, int r) { return sum(r) - sum(l - 1); }
  inline void init() { fill(c, c + maxn, 0); }
};
Bits tre[2];
int ql[maxn];
int qr[maxn];
class cmpr {
 public:
  bool operator()(const int& x, const int& y) const { return qr[x] < qr[y]; }
};
class cmpl {
 public:
  bool operator()(const int& x, const int& y) const { return ql[x] < ql[y]; }
};
int p[maxn];
long long ans[maxn];
int L[maxn];
int R[maxn];
int t;
int sk[maxn];
int id[maxn];
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int q;
  cin >> n >> q;
  p[0] = 2 * n;
  p[n + 1] = 2 * n;
  for (int i = (1); i <= (n); ++i) cin >> p[i];
  for (int i = (1); i <= (q); ++i) cin >> ql[i];
  for (int i = (1); i <= (q); ++i) cin >> qr[i], id[i] = i;
  t = -1;
  sk[++t] = 0;
  for (int i = (1); i <= (n); ++i) {
    while (p[i] > p[sk[t]]) t--;
    L[i] = sk[t];
    sk[++t] = i;
  }
  sk[t = 0] = n + 1;
  for (int i = (n); i >= (1); --i) {
    while (p[i] > p[sk[t]]) t--;
    R[i] = sk[t];
    sk[++t] = i;
  }
  sort(id + 1, id + 1 + q, cmpr());
  int pos = 1;
  for (int i = (1); i <= (q); ++i) {
    int x = id[i];
    while (pos <= qr[x]) {
      tre[0].modify(L[pos], L[pos]);
      tre[1].modify(L[pos], 1);
      pos++;
    }
    long long val = tre[0].qury(ql[x], qr[x]);
    long long cnt = tre[1].qury(ql[x], qr[x]);
    ans[x] -= val;
    ans[x] -= (qr[x] - ql[x] + 1 - cnt) * (ql[x] - 1);
  }
  tre[0].init();
  tre[1].init();
  sort(id + 1, id + 1 + q, cmpl());
  pos = n;
  for (int i = (q); i >= (1); --i) {
    int x = id[i];
    while (pos >= ql[x]) {
      tre[0].modify(R[pos], R[pos]);
      tre[1].modify(R[pos], 1);
      pos--;
    }
    long long val = tre[0].qury(ql[x], qr[x]);
    long long cnt = tre[1].qury(ql[x], qr[x]);
    ans[x] += val;
    ans[x] += (qr[x] - ql[x] + 1 - cnt) * (qr[x] + 1);
  }
  for (int i = (1); i <= (q); ++i) {
    if (i != 1) cout << " ";
    cout << ans[i] - (qr[i] - ql[i] + 1);
  }
  return 0;
}