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 NMAX = 1000100;
int l[NMAX], r[NMAX], a[NMAX];
long long ans[NMAX];
vector<int> lh[NMAX], rh[NMAX];
long long n, q;
struct segtree {
  vector<long long> t, d;
  int h, n;
  segtree() {}
  segtree(int sz) {
    h = 32 - __builtin_clz(sz);
    n = 1 << h;
    t = vector<long long>(n << 1, 0);
    d = vector<long long>(n << 1, 0);
  }
  void apply(int x, long long v) {
    d[x] += v;
    t[x] += v * (1 << (h - 31 + __builtin_clz(x)));
  }
  void pushh(int x) {
    if (d[x]) {
      apply(x << 1, d[x]);
      apply(x << 1 | 1, d[x]);
      d[x] = 0;
    }
  }
  void push(int i) {
    i += n;
    for (int l = h; l > 0; --l) {
      pushh(i >> l);
    }
  }
  void pulll(int x) {
    t[x] =
        t[x << 1] + t[x << 1 | 1] + d[x] * (1 << (h - 31 + __builtin_clz(x)));
  }
  void set(int l, int r) {
    if (l >= r) return;
    push(l);
    push(r - 1);
    bool cl = 0, cr = 0;
    for (l += n, r += n; l < r; l /= 2, r /= 2) {
      if (cl) {
        pulll(l - 1);
      }
      if (cr) {
        pulll(r);
      }
      if (l % 2) {
        apply(l++, 1);
        cl = 1;
      }
      if (r % 2) {
        apply(--r, 1);
        cr = 1;
      }
    }
    for (--l; r > 0; l /= 2, r /= 2) {
      if (cl) pulll(l);
      if (cr && (!cl || r != l)) pulll(r);
    }
  }
  long long query(int l, int r) {
    push(l);
    push(r - 1);
    long long ans = 0;
    for (l += n, r += n; l < r; l /= 2, r /= 2) {
      if (l % 2) {
        ans += t[l++];
      }
      if (r % 2) {
        ans += t[--r];
      }
    }
    return ans;
  }
};
inline char get() {
  static char buf[100000], *S = buf, *T = buf;
  if (S == T) {
    T = (S = buf) + fread(buf, 1, 100000, stdin);
    if (S == T) return EOF;
  }
  return *S++;
}
template <typename T>
inline void read(T& x) {
  static char c;
  x = 0;
  for (c = get(); c < '0' || c > '9'; c = get())
    ;
  for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  read(n);
  read(q);
  for (int i = 1; i <= n; ++i) {
    read(a[i]);
  }
  for (int i = 1; i <= q; ++i) {
    read(l[i]);
    lh[l[i]].push_back(i);
  }
  for (int i = 1; i <= q; ++i) {
    read(r[i]);
    rh[r[i]].push_back(i);
  }
  segtree st(n + 2);
  a[0] = NMAX;
  stack<int> s;
  s.push(0);
  for (int i = 1; i <= n; ++i) {
    while (!s.empty() && a[s.top()] < a[i]) s.pop();
    st.set(s.top() + 1, i);
    for (auto& j : rh[i]) {
      ans[j] += st.query(l[j], i);
    }
    s.push(i);
  }
  st = segtree(n + 1);
  s.pop();
  a[n + 1] = NMAX;
  s.push(n + 1);
  for (int i = n; i > 0; --i) {
    while (!s.empty() && a[s.top()] < a[i]) s.pop();
    st.set(i + 1, s.top());
    for (auto& j : lh[i]) {
      ans[j] += st.query(i, r[j] + 1);
    }
    s.push(i);
  }
  for (int i = 1; i <= q; ++i) {
    cout << ans[i] + r[i] - l[i] + 1 << " ";
  }
  cout << endl;
  return 0;
}