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;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
T div_floor(T a, T b) {
  if (b < 0) a *= -1, b *= -1;
  return a >= 0 ? a / b : (a + 1) / b - 1;
}
template <class T>
T div_ceil(T a, T b) {
  if (b < 0) a *= -1, b *= -1;
  return a > 0 ? (a - 1) / b + 1 : a / b;
}
constexpr long long mod = 1000000007;
constexpr long long INF = mod * mod;
constexpr int MAX = 200010;
namespace atcoder {
namespace internal {
int ceil_pow2(int n) {
  int x = 0;
  while ((1U << x) < (unsigned int)(n)) x++;
  return x;
}
int bsf(unsigned int n) { return __builtin_ctz(n); }
}  // namespace internal
}  // namespace atcoder
namespace atcoder {
template <class S, S (*op)(S, S), S (*e)()>
struct segtree {
 public:
  segtree() : segtree(0) {}
  segtree(int n) : segtree(std::vector<S>(n, e())) {}
  segtree(const std::vector<S> &v) : _n(int(v.size())) {
    log = internal::ceil_pow2(_n);
    size = 1 << log;
    d = std::vector<S>(2 * size, e());
    for (int i = 0; i < _n; i++) d[size + i] = v[i];
    for (int i = size - 1; i >= 1; i--) {
      update(i);
    }
  }
  void set(int p, S x) {
    assert(0 <= p && p < _n);
    p += size;
    d[p] = x;
    for (int i = 1; i <= log; i++) update(p >> i);
  }
  S get(int p) {
    assert(0 <= p && p < _n);
    return d[p + size];
  }
  S prod(int l, int r) {
    assert(0 <= l && l <= r && r <= _n);
    S sml = e(), smr = e();
    l += size;
    r += size;
    while (l < r) {
      if (l & 1) sml = op(sml, d[l++]);
      if (r & 1) smr = op(d[--r], smr);
      l >>= 1;
      r >>= 1;
    }
    return op(sml, smr);
  }
  S all_prod() { return d[1]; }
  template <bool (*f)(S)>
  int max_right(int l) {
    return max_right(l, [](S x) { return f(x); });
  }
  template <class F>
  int max_right(int l, F f) {
    assert(0 <= l && l <= _n);
    assert(f(e()));
    if (l == _n) return _n;
    l += size;
    S sm = e();
    do {
      while (l % 2 == 0) l >>= 1;
      if (!f(op(sm, d[l]))) {
        while (l < size) {
          l = (2 * l);
          if (f(op(sm, d[l]))) {
            sm = op(sm, d[l]);
            l++;
          }
        }
        return l - size;
      }
      sm = op(sm, d[l]);
      l++;
    } while ((l & -l) != l);
    return _n;
  }
  template <bool (*f)(S)>
  int min_left(int r) {
    return min_left(r, [](S x) { return f(x); });
  }
  template <class F>
  int min_left(int r, F f) {
    assert(0 <= r && r <= _n);
    assert(f(e()));
    if (r == 0) return 0;
    r += size;
    S sm = e();
    do {
      r--;
      while (r > 1 && (r % 2)) r >>= 1;
      if (!f(op(d[r], sm))) {
        while (r < size) {
          r = (2 * r + 1);
          if (f(op(d[r], sm))) {
            sm = op(d[r], sm);
            r--;
          }
        }
        return r + 1 - size;
      }
      sm = op(d[r], sm);
    } while ((r & -r) != r);
    return 0;
  }

 private:
  int _n, size, log;
  std::vector<S> d;
  void update(int k) { d[k] = op(d[2 * k], d[2 * k + 1]); }
};
}  // namespace atcoder
namespace atcoder {
template <class S, S (*op)(S, S), S (*e)(), class F, S (*mapping)(F, S),
          F (*composition)(F, F), F (*id)()>
struct lazy_segtree {
 public:
  lazy_segtree() : lazy_segtree(0) {}
  lazy_segtree(int n) : lazy_segtree(std::vector<S>(n, e())) {}
  lazy_segtree(const std::vector<S> &v) : _n(int(v.size())) {
    log = internal::ceil_pow2(_n);
    size = 1 << log;
    d = std::vector<S>(2 * size, e());
    lz = std::vector<F>(size, id());
    for (int i = 0; i < _n; i++) d[size + i] = v[i];
    for (int i = size - 1; i >= 1; i--) {
      update(i);
    }
  }
  void set(int p, S x) {
    assert(0 <= p && p < _n);
    p += size;
    for (int i = log; i >= 1; i--) push(p >> i);
    d[p] = x;
    for (int i = 1; i <= log; i++) update(p >> i);
  }
  S get(int p) {
    assert(0 <= p && p < _n);
    p += size;
    for (int i = log; i >= 1; i--) push(p >> i);
    return d[p];
  }
  S prod(int l, int r) {
    assert(0 <= l && l <= r && r <= _n);
    if (l == r) return e();
    l += size;
    r += size;
    for (int i = log; i >= 1; i--) {
      if (((l >> i) << i) != l) push(l >> i);
      if (((r >> i) << i) != r) push(r >> i);
    }
    S sml = e(), smr = e();
    while (l < r) {
      if (l & 1) sml = op(sml, d[l++]);
      if (r & 1) smr = op(d[--r], smr);
      l >>= 1;
      r >>= 1;
    }
    return op(sml, smr);
  }
  S all_prod() { return d[1]; }
  void apply(int p, F f) {
    assert(0 <= p && p < _n);
    p += size;
    for (int i = log; i >= 1; i--) push(p >> i);
    d[p] = mapping(f, d[p]);
    for (int i = 1; i <= log; i++) update(p >> i);
  }
  void apply(int l, int r, F f) {
    assert(0 <= l && l <= r && r <= _n);
    if (l == r) return;
    l += size;
    r += size;
    for (int i = log; i >= 1; i--) {
      if (((l >> i) << i) != l) push(l >> i);
      if (((r >> i) << i) != r) push((r - 1) >> i);
    }
    {
      int l2 = l, r2 = r;
      while (l < r) {
        if (l & 1) all_apply(l++, f);
        if (r & 1) all_apply(--r, f);
        l >>= 1;
        r >>= 1;
      }
      l = l2;
      r = r2;
    }
    for (int i = 1; i <= log; i++) {
      if (((l >> i) << i) != l) update(l >> i);
      if (((r >> i) << i) != r) update((r - 1) >> i);
    }
  }
  template <bool (*g)(S)>
  int max_right(int l) {
    return max_right(l, [](S x) { return g(x); });
  }
  template <class G>
  int max_right(int l, G g) {
    assert(0 <= l && l <= _n);
    assert(g(e()));
    if (l == _n) return _n;
    l += size;
    for (int i = log; i >= 1; i--) push(l >> i);
    S sm = e();
    do {
      while (l % 2 == 0) l >>= 1;
      if (!g(op(sm, d[l]))) {
        while (l < size) {
          push(l);
          l = (2 * l);
          if (g(op(sm, d[l]))) {
            sm = op(sm, d[l]);
            l++;
          }
        }
        return l - size;
      }
      sm = op(sm, d[l]);
      l++;
    } while ((l & -l) != l);
    return _n;
  }
  template <bool (*g)(S)>
  int min_left(int r) {
    return min_left(r, [](S x) { return g(x); });
  }
  template <class G>
  int min_left(int r, G g) {
    assert(0 <= r && r <= _n);
    assert(g(e()));
    if (r == 0) return 0;
    r += size;
    for (int i = log; i >= 1; i--) push((r - 1) >> i);
    S sm = e();
    do {
      r--;
      while (r > 1 && (r % 2)) r >>= 1;
      if (!g(op(d[r], sm))) {
        while (r < size) {
          push(r);
          r = (2 * r + 1);
          if (g(op(d[r], sm))) {
            sm = op(d[r], sm);
            r--;
          }
        }
        return r + 1 - size;
      }
      sm = op(d[r], sm);
    } while ((r & -r) != r);
    return 0;
  }

 private:
  int _n, size, log;
  std::vector<S> d;
  std::vector<F> lz;
  void update(int k) { d[k] = op(d[2 * k], d[2 * k + 1]); }
  void all_apply(int k, F f) {
    d[k] = mapping(f, d[k]);
    if (k < size) lz[k] = composition(f, lz[k]);
  }
  void push(int k) {
    all_apply(2 * k, lz[k]);
    all_apply(2 * k + 1, lz[k]);
    lz[k] = id();
  }
};
}  // namespace atcoder
using namespace atcoder;
int target;
struct Max {
  int val;
  static Max op(Max a, Max b) { return {max(a.val, b.val)}; }
  static Max e() { return {0}; }
  static bool f(Max x) { return x.val <= target; }
};
struct Sum {
  long long val, len;
  static Sum op(Sum a, Sum b) { return {a.val + b.val, a.len + b.len}; }
  static Sum e() { return {0, 0}; }
};
struct Add {
  long long val;
  static Sum mapping(Add f, Sum x) { return {x.val + f.val * x.len, x.len}; }
  static Add composition(Add f, Add g) { return {f.val + g.val}; }
  static Add id() { return {0}; }
};
struct query {
  int l, r, idx;
};
void solve(vector<int> &a, vector<query> &qs, vector<long long> &ans) {
  int n = a.size();
  vector<Max> a_max(n);
  for (int((i)) = (int)(0); ((i)) < (int)((n)); ++((i))) a_max[i] = {a[i]};
  segtree<Max, Max::op, Max::e> st_max(a_max);
  lazy_segtree<Sum, Sum::op, Sum::e, Add, Add::mapping, Add::composition,
               Add::id>
      st_sum(vector<Sum>(n, {0, 1}));
  sort(qs.begin(), qs.end(),
       [](const query &q1, const query &q2) { return q1.r < q2.r; });
  int cur = 0;
  for (auto &q : qs) {
    while (cur < q.r) {
      target = a[cur];
      int l = st_max.min_left<Max::f>(cur + 1);
      st_sum.apply(l, cur + 1, {1});
      ++cur;
    }
    ans[q.idx] += st_sum.prod(q.l, n).val;
  }
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  vector<int> a(n);
  for (int((i)) = (int)(0); ((i)) < (int)((n)); ++((i))) scanf("%d", &a[i]);
  int ls[q], rs[q];
  for (int((i)) = (int)(0); ((i)) < (int)((q)); ++((i)))
    scanf("%d", &ls[i]), --ls[i];
  for (int((i)) = (int)(0); ((i)) < (int)((q)); ++((i))) scanf("%d", &rs[i]);
  vector<long long> ans(q, 0);
  {
    vector<query> qs;
    for (int((i)) = (int)(0); ((i)) < (int)((q)); ++((i)))
      qs.push_back({ls[i], rs[i], i});
    solve(a, qs, ans);
  }
  {
    vector<query> qs;
    for (int((i)) = (int)(0); ((i)) < (int)((q)); ++((i)))
      qs.push_back({n - rs[i], n - ls[i], i});
    reverse(a.begin(), a.end());
    solve(a, qs, ans);
  }
  for (int((i)) = (int)(0); ((i)) < (int)((q)); ++((i)))
    ans[i] -= rs[i] - ls[i];
  for (auto x : ans) printf("%lld ", x);
  printf("\n");
}