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 mod = 1e9 + 7;
const int base = 19260817;
const long long INF = 1e18;
const int maxn = 1e6 + 10;
const double eps = 1e-8;
int add(int x, int y) {
  x += y;
  return x >= mod ? x - mod : x;
}
int sub(int x, int y) {
  x -= y;
  return x < 0 ? x + mod : x;
}
int mul(int x, int y) { return 1LL * x * y % mod; }
int p[maxn], L[maxn], R[maxn];
long long c[maxn], d[maxn], ans[maxn];
pair<int, int> que[maxn];
stack<int> s;
vector<int> st[maxn], ed[maxn], qs[maxn], qe[maxn];
int lowbit(int o) { return o & (-o); }
void update(long long *tr, int o, int v) {
  while (o < maxn) {
    tr[o] += v;
    o += lowbit(o);
  }
}
long long query(long long *tr, int o) {
  long long res = 0;
  while (o) {
    res += tr[o];
    o -= lowbit(o);
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> p[i];
  for (int i = 1; i <= q; i++) cin >> que[i].first;
  for (int i = 1; i <= q; i++) cin >> que[i].second;
  for (int i = 1; i <= n; i++) {
    while (!s.empty() && p[s.top()] < p[i]) R[s.top()] = i - 1, s.pop();
    s.push(i);
  }
  while (!s.empty()) R[s.top()] = n, s.pop();
  for (int i = n; i >= 1; i--) {
    while (!s.empty() && p[s.top()] < p[i]) L[s.top()] = i + 1, s.pop();
    s.push(i);
  }
  while (!s.empty()) L[s.top()] = 1, s.pop();
  for (int i = 1; i <= n; i++) st[L[i]].push_back(i), ed[R[i]].push_back(i);
  for (int i = 1; i <= q; i++)
    qs[que[i].first].push_back(i), qe[que[i].second].push_back(i);
  for (int i = 1; i <= n; i++) update(c, i, L[i]);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < qs[i].size(); j++) {
      int id = qs[i][j];
      ans[id] -= query(c, que[id].second) - query(c, que[id].first - 1);
      ans[id] -= (query(d, que[id].second) - query(d, que[id].first - 1)) * i;
    }
    for (int j = 0; j < st[i].size(); j++) {
      int id = st[i][j];
      update(c, id, -L[id]);
      update(d, id, 1);
    }
  }
  memset(c, 0, sizeof(c)), memset(d, 0, sizeof(d));
  for (int i = 1; i <= n; i++) update(c, i, R[i]);
  for (int i = n; i >= 1; i--) {
    for (int j = 0; j < qe[i].size(); j++) {
      int id = qe[i][j];
      ans[id] += query(c, que[id].second) - query(c, que[id].first - 1);
      ans[id] += (query(d, que[id].second) - query(d, que[id].first - 1)) * i;
    }
    for (int j = 0; j < ed[i].size(); j++) {
      int id = ed[i][j];
      update(c, id, -R[id]);
      update(d, id, 1);
    }
  }
  for (int i = 1; i <= q; i++)
    cout << ans[i] + que[i].second - que[i].first + 1 << " ";
  return 0;
}