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;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  return x * f;
}
const long long MAXN = 2000010;
const long long INF = 1e9;
long long N, Q;
long long s[MAXN << 2], tag[MAXN << 2];
long long sta[MAXN + 1], top;
long long R[MAXN + 1], L[MAXN + 1];
long long ans[MAXN + 1];
long long a[MAXN + 1];
struct data {
  long long l, r, id;
} q[MAXN + 1];
inline void tage_lazy(long long rt, long long l, long long r) {
  if (tag[rt]) {
    long long mid = (l + r) >> 1;
    tag[rt << 1] += tag[rt];
    tag[rt << 1 | 1] += tag[rt];
    s[rt << 1] += (mid - l + 1) * tag[rt];
    s[rt << 1 | 1] += (r - mid) * tag[rt];
    tag[rt] = 0;
  }
  return;
}
inline void update(long long rt) {
  s[rt] = s[rt << 1] + s[rt << 1 | 1];
  return;
}
inline long long Query(long long rt, long long l, long long r, long long L,
                       long long R) {
  if (L <= l && R >= r) return s[rt];
  long long mid = (l + r) >> 1, Ans = 0;
  tage_lazy(rt, l, r);
  if (L <= mid) Ans += Query(rt << 1, l, mid, L, R);
  if (R > mid) Ans += Query(rt << 1 | 1, mid + 1, r, L, R);
  return Ans;
}
inline void Add(long long rt, long long l, long long r, long long L,
                long long R, long long d) {
  if (L <= l && R >= r) {
    s[rt] += d * (r - l + 1);
    tag[rt] += d;
    return;
  }
  tage_lazy(rt, l, r);
  long long mid = (l + r) >> 1;
  if (L <= mid) Add(rt << 1, l, mid, L, R, d);
  if (R > mid) Add(rt << 1 | 1, mid + 1, r, L, R, d);
  update(rt);
  return;
}
bool cmp(data a, data b) { return a.l < b.l; }
inline void solve() {
  top = 0;
  memset(s, 0, sizeof(s));
  memset(tag, 0, sizeof(tag));
  for (long long i = 1; i <= N; i++) {
    L[i] = i;
    while (top && a[sta[top]] < a[i]) R[sta[top]] = i - 1, --top;
    sta[++top] = i;
  }
  for (long long i = 1; i <= top; i++) R[sta[i]] = N;
  long long nw = Q;
  sort(q + 1, q + Q + 1, cmp);
  for (long long i = N; i >= 1; i--) {
    Add(1, 1, N, L[i], R[i], 1);
    while (q[nw].l == i && nw >= 1)
      ans[q[nw].id] += Query(1, 1, N, i, q[nw].r), --nw;
  }
  return;
}
int main() {
  N = read(), Q = read();
  for (long long i = 1; i <= N; i++) a[i] = read();
  for (long long i = 1; i <= Q; i++) q[i].l = read(), q[i].id = i;
  for (long long i = 1; i <= Q; i++)
    q[i].r = read(), ans[i] -= (q[i].r - q[i].l + 1);
  solve();
  reverse(a + 1, a + N + 1);
  for (long long i = 1; i <= Q; i++)
    q[i].l = N - q[i].l + 1, q[i].r = N - q[i].r + 1, swap(q[i].l, q[i].r);
  solve();
  for (long long i = 1; i <= Q; i++) printf("%lld ", ans[i]);
  return 0;
}