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("Ofast")
using namespace std;
const long long MAXN = 1000005, INF = 1e9;
struct mode {
  long long first, second, i;
} query[MAXN];
long long num[MAXN], lf[MAXN], rg[MAXN], ans[MAXN];
long long bit[2][MAXN], n, flag;
stack<long long> st;
bool cmp1(mode a, mode b) { return a.second < b.second; }
bool cmp2(mode a, mode b) { return a.first > b.first; }
void modify(long long x, long long v, long long t) {
  if (flag == 1) x = n - x + 1;
  for (; x <= n; x += x & -x) bit[t][x] += v;
}
long long Query(long long x, long long t) {
  long long re = 0;
  if (flag == 1) x = n - x + 1;
  for (; x; x -= x & -x) re += bit[t][x];
  return re;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long q, t;
  cin >> n >> q, num[0] = num[n + 1] = INF;
  for (int i = 1; i <= n; ++i) cin >> num[i];
  for (int i = 1; i <= q; ++i) cin >> query[i].first, query[i].i = i;
  for (int i = 1; i <= q; ++i)
    cin >> query[i].second, ans[i] += query[i].second - query[i].first + 1;
  sort(query + 1, query + q + 1, cmp1), st.push(0), t = 1;
  for (int i = 1; i <= n; ++i) {
    while (num[i] > num[st.top()]) st.pop();
    lf[i] = st.top(), st.push(i);
    if (lf[i])
      modify(1, i - lf[i] - 1, 0), modify(lf[i] + 1, -(i - lf[i] - 1), 0);
    modify(lf[i] + 1, i, 0), modify(i + 1, -i, 0), modify(lf[i] + 1, 1, 1),
        modify(i + 1, -1, 1);
    while (t <= q && query[t].second <= i)
      ans[query[t].i] +=
          Query(query[t].first, 0) - Query(query[t].first, 1) * query[t].first,
          ++t;
  }
  while (!st.empty()) st.pop();
  memset(bit, 0, sizeof bit), sort(query + 1, query + q + 1, cmp2),
      st.push(n + 1), t = 1, flag = 1;
  for (int i = n; i > 0; --i) {
    while (num[i] > num[st.top()]) st.pop();
    rg[i] = st.top(), st.push(i);
    if (rg[i] <= n)
      modify(n, rg[i] - i - 1, 0), modify(rg[i] - 1, -(rg[i] - i - 1), 0);
    modify(rg[i] - 1, -i, 0), modify(i - 1, i, 0), modify(rg[i] - 1, 1, 1),
        modify(i - 1, -1, 1);
    while (t <= q && query[t].first >= i)
      ans[query[t].i] += Query(query[t].second, 0) +
                         Query(query[t].second, 1) * query[t].second,
          ++t;
  }
  for (int i = 1; i <= q; ++i) cout << ans[i] << " \n"[i == q];
}