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 N = 1e6 + 7;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-8;
const double PI = acos(-1);
template <class T, class S>
inline void add(T& a, S b) {
  a += b;
  if (a >= mod) a -= mod;
}
template <class T, class S>
inline void sub(T& a, S b) {
  a -= b;
  if (a < 0) a += mod;
}
template <class T, class S>
inline bool chkmax(T& a, S b) {
  return a < b ? a = b, true : false;
}
template <class T, class S>
inline bool chkmin(T& a, S b) {
  return a > b ? a = b, true : false;
}
struct Bit {
  long long a[N];
  void init() { memset(a, 0, sizeof(a)); }
  void modify(int x, int v) {
    if (!x) return;
    for (int i = x; i < N; i += i & -i) a[i] += v;
  }
  long long sum(int x) {
    long long ans = 0;
    for (int i = x; i; i -= i & -i) ans += a[i];
    return ans;
  }
  long long query(int L, int R) { return sum(R) - sum(L - 1); }
} bit[2];
int n, q, ql[N], qr[N];
int a[N], L[N], R[N];
int id[N];
long long ans[N];
stack<int> stk;
vector<int> qus[N];
bool cmpr(const int& a, const int& b) { return qr[a] < qr[b]; }
bool cmpl(const int& a, const int& b) { return ql[a] < ql[b]; }
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  while (((int)stk.size())) stk.pop();
  for (int i = 1; i <= n; i++) {
    while (((int)stk.size()) && a[stk.top()] < a[i]) stk.pop();
    L[i] = ((int)stk.size()) ? stk.top() : 0;
    stk.push(i);
  }
  while (((int)stk.size())) stk.pop();
  for (int i = n; i >= 1; i--) {
    while (((int)stk.size()) && a[stk.top()] < a[i]) stk.pop();
    R[i] = ((int)stk.size()) ? stk.top() : n + 1;
    stk.push(i);
  }
  for (int i = 1; i <= q; i++) scanf("%d", &ql[i]), id[i] = i;
  for (int i = 1; i <= q; i++) scanf("%d", &qr[i]);
  sort(id + 1, id + 1 + q, cmpr);
  for (int i = 1, j = 1; i <= q; i++) {
    int x = id[i];
    while (j <= qr[x]) bit[0].modify(L[j], L[j]), bit[1].modify(L[j], 1), j++;
    long long val = bit[0].query(ql[x], qr[x]),
              cnt = bit[1].query(ql[x], qr[x]);
    ans[x] -= val + 1LL * (qr[x] - ql[x] + 1 - cnt) * (ql[x] - 1);
  }
  sort(id + 1, id + 1 + q, cmpl);
  bit[0].init();
  bit[1].init();
  for (int i = q, j = n; i >= 1; i--) {
    int x = id[i];
    while (j >= ql[x]) bit[0].modify(R[j], R[j]), bit[1].modify(R[j], 1), j--;
    long long val = bit[0].query(ql[x], qr[x]),
              cnt = bit[1].query(ql[x], qr[x]);
    ans[x] += val + 1LL * (qr[x] - ql[x] + 1 - cnt) * (qr[x] + 1);
  }
  for (int i = 1; i <= q; i++)
    printf("%lld%c", ans[i] - (qr[i] - ql[i] + 1), " \n"[i == q]);
  return 0;
}