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>
#pragma GCC optimize("O2")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2,fma,tune=native")
using namespace std;
int Pow(int a, int b, int md, int ans = 1) {
  for (; b; b >>= 1, a = a * a % md)
    if (b & 1) ans = ans * a % md;
  return ans % md;
}
const int MAXN = 1e6 + 10;
const int INF = 1e9 + 7;
const int MOD = 1e9 + 7;
int lazy[MAXN << 2], A[MAXN], Ql[MAXN], Qr[MAXN], L[MAXN], R[MAXN], n, q;
long long int ans[MAXN], seg[MAXN << 2];
vector<int> Q[2][MAXN];
stack<int> st;
void shift(int id, int l, int r) {
  seg[id] += 1LL * lazy[id] * (r - l);
  if (r - l > 1) {
    lazy[id << 1 | 1] += lazy[id];
    lazy[id << 1] += lazy[id];
  }
  lazy[id] = 0;
}
void update(int ql, int qr, int id = 1, int l = 0, int r = MAXN) {
  shift(id, l, r);
  if (qr <= l || r <= ql || ql >= qr) return;
  if (ql <= l && r <= qr) {
    lazy[id]++;
    shift(id, l, r);
    return;
  }
  int mid = (l + r) >> 1;
  update(ql, qr, id << 1, l, mid);
  update(ql, qr, id << 1 | 1, mid, r);
  seg[id] = seg[id << 1] + seg[id << 1 | 1];
}
long long int get(int ql, int qr, int id = 1, int l = 0, int r = MAXN) {
  shift(id, l, r);
  if (qr <= l || r <= ql || ql >= qr) return 0;
  if (ql <= l && r <= qr) return seg[id];
  int mid = (l + r) >> 1;
  return get(ql, qr, id << 1, l, mid) + get(ql, qr, id << 1 | 1, mid, r);
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &A[i]);
  }
  A[0] = A[n + 1] = INF;
  st.push(0);
  for (int i = 1; i <= n; i++) {
    while (A[st.top()] < A[i]) st.pop();
    L[i] = st.top();
    st.push(i);
  }
  while ((int)st.size()) st.pop();
  st.push(n + 1);
  for (int i = n; i; i--) {
    while (A[st.top()] < A[i]) st.pop();
    R[i] = st.top();
    st.push(i);
  }
  for (int i = 1; i <= q; i++) {
    scanf("%d", &Ql[i]);
    Q[1][Ql[i]].push_back(i);
  }
  for (int i = 1; i <= q; i++) {
    scanf("%d", &Qr[i]);
    Q[0][Qr[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    update(L[i] + 1, i);
    for (int id : Q[0][i]) ans[id] += get(Ql[id], Qr[id] + 1);
  }
  memset(seg, 0, sizeof seg);
  memset(lazy, 0, sizeof lazy);
  for (int i = n; i; i--) {
    update(i, R[i]);
    for (int id : Q[1][i]) ans[id] += get(Ql[id], Qr[id] + 1);
  }
  for (int i = 1; i <= q; i++) printf("%lld ", ans[i]);
  return 0;
}