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 long long N = 1e6 + 7;
long long n, q;
long long a[N];
long long l[N], r[N];
void read() {
  cin >> n >> q;
  for (long long i = 0; i < n; ++i) cin >> a[i];
  for (long long i = 0; i < q; ++i) cin >> l[i];
  for (long long i = 0; i < q; ++i) cin >> r[i];
  for (long long i = 0; i < q; ++i) {
    --l[i];
    --r[i];
  }
}
long long ans[N];
long long link[N];
vector<long long> add[N], del[N];
long long f[N], fcnt[N];
void addf(long long i) {
  long long t = i;
  for (; i < N; i |= (i + 1)) {
    f[i] += t;
  }
}
void addfcnt(long long i) {
  for (; i < N; i |= (i + 1)) {
    fcnt[i]++;
  }
}
void add_(long long i) {
  addf(i);
  addfcnt(i);
}
long long getsum(long long i) {
  long long ans = 0;
  for (; i >= 0; i = (i & (i + 1)) - 1) {
    ans += f[i];
  }
  return ans;
}
long long getcnt(long long i) {
  long long ans = 0;
  for (; i >= 0; i = (i & (i + 1)) - 1) {
    ans += fcnt[i];
  }
  return ans;
}
long long getmax(long long x) {
  return getcnt(x) * x + getsum(N - 1) - getsum(x);
}
long long pref[N];
void kek() {
  for (long long i = 0; i < N; ++i) {
    f[i] = fcnt[i] = 0;
  }
  for (long long i = 0; i < q; ++i) {
    ans[i] += pref[r[i] + 1] - pref[l[i]];
  }
  for (long long i = 0; i < N; ++i) f[i] = fcnt[i] = 0;
  for (long long i = 0; i < n; ++i) {
    add[i].clear();
    del[i].clear();
  }
  for (long long i = 0; i < n; ++i) link[i] = -1;
  vector<long long> st;
  for (long long i = n - 1; i >= 0; --i) {
    while (st.size() && a[i] > a[st.back()]) {
      link[st.back()] = i;
      st.pop_back();
    }
    st.push_back(i);
  }
  for (long long i = 0; i < q; ++i) {
    add[r[i]].push_back(i);
    if (l[i]) del[l[i] - 1].push_back(i);
  }
  for (long long i = 0; i < n; ++i) {
    add_(link[i] + 1);
    for (long long e : add[i]) {
      ans[e] -= getmax(l[e]);
    }
    for (long long e : del[i]) {
      ans[e] += getmax(l[e]);
    }
  }
}
void solve() {
  for (long long i = 0; i < n; ++i) {
    pref[i + 1] = pref[i] + i;
  }
  for (long long i = 0; i < q; ++i) {
    ans[i] = r[i] - l[i] + 1;
  }
  kek();
  reverse(a, a + n);
  for (long long i = 0; i < q; ++i) {
    l[i] = n - 1 - l[i];
    r[i] = n - 1 - r[i];
    swap(l[i], r[i]);
  }
  kek();
}
void print() {
  for (long long i = 0; i < q; ++i) cout << ans[i] << ' ';
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(20);
  read();
  solve();
  print();
}