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;
using INT = long long;
using pii = pair<int, int>;
const int NN = 2727272;
int a[NN], pre[NN], nxt[NN];
int ll[NN], rr[NN];
int L[NN], R[NN], MID[NN];
vector<pii> Q[NN], QQ[NN];
void build(int u, int l, int r) {
  L[u] = l;
  R[u] = r;
  if (l + 1 == r) return;
  MID[u] = (l + r >> 1);
  build((u << 1), l, (l + r >> 1));
  build(((u << 1) | 1), (l + r >> 1), r);
}
INT s[NN], fl[NN], ans[NN];
int st, ed;
void push_down(int u) {
  if (!fl[u]) return;
  fl[(u << 1)] += fl[u];
  fl[((u << 1) | 1)] += fl[u];
  s[(u << 1)] += fl[u] * (MID[u] - L[u]);
  s[((u << 1) | 1)] += fl[u] * (R[u] - MID[u]);
  fl[u] = 0;
}
void add(int u) {
  if (R[u] <= st or ed <= L[u]) return;
  s[u] += min(ed, R[u]) - max(L[u], st);
  if (st <= L[u] and R[u] <= ed) {
    fl[u]++;
    return;
  }
  push_down(u);
  add((u << 1));
  add(((u << 1) | 1));
}
INT calc(int u) {
  if (R[u] <= st or ed <= L[u]) return 0;
  if (st <= L[u] and R[u] <= ed) return s[u];
  push_down(u);
  return calc((u << 1)) + calc(((u << 1) | 1));
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  build(1, 1, n + 1);
  for (int i = 1; i <= n; i++) {
    int u = i - 1;
    while (u and a[u] <= a[i]) u = pre[u];
    pre[i] = u;
  }
  nxt[n + 1] = n + 1;
  for (int i = n; i; i--) {
    int u = i + 1;
    while (u <= n and a[u] <= a[i]) u = nxt[u];
    nxt[i] = u;
  }
  for (int i = 1; i <= q; i++) scanf("%d", ll + i);
  for (int i = 1; i <= q; i++)
    scanf("%d", rr + i), Q[rr[i]].push_back({ll[i], i}),
        QQ[ll[i] - 1].push_back({rr[i], i});
  for (int i = 1; i <= n; i++) {
    st = pre[i] + 1;
    ed = nxt[i];
    add(1);
    for (pii p : QQ[i]) st = i + 1, ed = p.first + 1, ans[p.second] -= calc(1);
    for (pii p : Q[i]) st = p.first, ed = i + 1, ans[p.second] += calc(1);
  }
  for (int i = 1; i <= q; i++) printf("%I64d ", ans[i]);
  return 0;
}