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;
const int maxn = 1000010;
int n, a[maxn], Q, S[maxn], top, L[maxn], R[maxn], cnt;
long long ans[maxn], c1[maxn], c2[maxn];
struct node {
  int l, r;
} e[maxn];
struct node2 {
  int x, l, r, id;
} p[maxn * 2];
bool cmp1(node2 a, node2 b) {
  if (a.x == b.x) return a.id < b.id;
  return a.x > b.x;
}
bool cmp2(node2 a, node2 b) {
  if (a.x == b.x) return a.id < b.id;
  return a.x < b.x;
}
void update(long long *c, int x, long long v) {
  while (x <= n) {
    c[x] += v;
    x += x & (-x);
  }
}
long long query(long long *c, int x) {
  long long res = 0;
  while (x) {
    res += c[x];
    x -= x & (-x);
  }
  return res;
}
void solve2() {
  memset(c1, 0, sizeof(c1));
  memset(c2, 0, sizeof(c2));
  cnt = 0;
  for (int i = 1; i <= n; i++) {
    ++cnt;
    p[cnt].x = R[i];
    p[cnt].l = i;
    p[cnt].id = 0;
  }
  for (int i = 1; i <= Q; i++) {
    ++cnt;
    p[cnt].x = e[i].r;
    p[cnt].l = e[i].l;
    p[cnt].r = e[i].r;
    p[cnt].id = i;
  }
  sort(p + 1, p + 1 + cnt, cmp2);
  for (int i = 1; i <= cnt; i++) {
    if (p[i].id == 0) {
      update(c1, p[i].l, 1);
      update(c2, p[i].l, p[i].x);
    } else {
      long long temp1 = query(c1, p[i].r) - query(c1, p[i].l - 1);
      long long temp2 = query(c2, p[i].r) - query(c2, p[i].l - 1);
      ans[p[i].id] += temp2;
      ans[p[i].id] += 1LL * (p[i].r - p[i].l + 1 - temp1) * p[i].x;
    }
  }
}
void solve1() {
  memset(c1, 0, sizeof(c1));
  memset(c2, 0, sizeof(c2));
  cnt = 0;
  for (int i = 1; i <= n; i++) {
    ++cnt;
    p[cnt].x = L[i];
    p[cnt].l = i;
    p[cnt].id = 0;
  }
  for (int i = 1; i <= Q; i++) {
    ++cnt;
    p[cnt].x = e[i].l;
    p[cnt].l = e[i].l;
    p[cnt].r = e[i].r;
    p[cnt].id = i;
  }
  sort(p + 1, p + 1 + cnt, cmp1);
  for (int i = 1; i <= cnt; i++) {
    if (p[i].id == 0) {
      update(c1, p[i].l, 1);
      update(c2, p[i].l, p[i].x);
    } else {
      long long temp1 = query(c1, p[i].r) - query(c1, p[i].l - 1);
      long long temp2 = query(c2, p[i].r) - query(c2, p[i].l - 1);
      ans[p[i].id] -= temp2;
      ans[p[i].id] -= 1LL * (p[i].r - p[i].l + 1 - temp1) * p[i].x;
    }
  }
}
int main() {
  scanf("%d%d", &n, &Q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= Q; i++) scanf("%d", &e[i].l);
  for (int i = 1; i <= Q; i++) scanf("%d", &e[i].r);
  for (int i = 1; i <= n; i++) {
    while (top && a[S[top]] < a[i]) top--;
    L[i] = S[top] + 1;
    S[++top] = i;
  }
  memset(S, 0, sizeof(S));
  top = 0;
  S[0] = n + 1;
  for (int i = n; i >= 1; i--) {
    while (top && a[S[top]] < a[i]) top--;
    R[i] = S[top] - 1;
    S[++top] = i;
  }
  solve1();
  solve2();
  for (int i = 1; i <= Q; i++) printf("%lld ", e[i].r - e[i].l + 1 + ans[i]);
  return 0;
}