Ilya is sitting in a waiting area of Metropolis airport and is bored of looking at time table that shows again and again that his plane is delayed. So he took out a sheet of paper and decided to solve some problems.
First Ilya has drawn a grid of size n × n and marked n squares on it, such that no two marked squares share the same row or the same column. He calls a rectangle on a grid with sides parallel to grid sides beautiful if exactly two of its corner squares are marked. There are exactly n·(n - 1) / 2 beautiful rectangles.
Ilya has chosen q query rectangles on a grid with sides parallel to grid sides (not necessarily beautiful ones), and for each of those rectangles he wants to find its beauty degree. Beauty degree of a rectangle is the number of beautiful rectangles that share at least one square with the given one.
Now Ilya thinks that he might not have enough time to solve the problem till the departure of his flight. You are given the description of marked cells and the query rectangles, help Ilya find the beauty degree of each of the query rectangles.
Input
The first line of input contains two integers n and q (2 ≤ n ≤ 200 000, 1 ≤ q ≤ 200 000) — the size of the grid and the number of query rectangles.
The second line contains n integers p1, p2, …, pn, separated by spaces (1 ≤ pi ≤ n, all pi are different), they specify grid squares marked by Ilya: in column i he has marked a square at row pi, rows are numbered from 1 to n, bottom to top, columns are numbered from 1 to n, left to right.
The following q lines describe query rectangles. Each rectangle is described by four integers: l, d, r, u (1 ≤ l ≤ r ≤ n, 1 ≤ d ≤ u ≤ n), here l and r are the leftmost and the rightmost columns of the rectangle, d and u the bottommost and the topmost rows of the rectangle.
Output
For each query rectangle output its beauty degree on a separate line.
Examples
input
2 3
1 2
1 1 1 1
1 1 1 2
1 1 2 2
output
1
1
1
input
4 2
1 3 2 4
4 1 4 4
1 1 2 3
output
3
5
题目大意：在一个n行n列的矩阵中，有n个点，每一列有一个点，然后有m次询问，每一次询问中，给出一个矩形，问由矩阵中的n个点组成的矩形中有多少个矩形与给定的矩形是相交的。
解题思路：将矩形的四条边无限延长，然后就可以分成9个部分，然后就画图看出这些排列组合是我们所求的答案。我们求每个部分的点的数量可以使用树状数组维护，对于y排序，然后将x加入树状数组，我们就可以算出y小于等于一个值时，x小于等于一个值的答案啦。然后再容斥一下就可以算出每个部分的答案啦。
代码：
#pragma GCC optimize(2)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <bitset>
#include <queue>
using namespace std;
#define int long long
#define ls root<<1
#define rs root<<1|1
const int maxn = 5e5 + 7;
const int inf = 0x3f3f3f3f;
int a[maxn][11], f[maxn];
struct Node
{
int x, y;
bool operator <(Node a)const {
return y < a.y;
}
} node[maxn];
struct Edge
{
int x, y, pos, id;
bool operator <(Edge a)const{
return y < a.y;
}
} e[maxn << 2];
int sum[maxn], n, m;
void add(int x)
{
while (x <= n){
sum[x]++;
x += (x & -x);
}
}
int ask(int x)
{
int ans = 0;
while(x){
ans += sum[x];
x -= (x & -x);
}
return ans;
}
int cnt, high[10], tot;
signed main()
{
std::ios::sync_with_stdio(false);
cin.tie(0);
cin >> n >> m;
for (int i = 1; i <= n;i++){
int y;
cin >> y;
node[i] = Node{i, y};
}
for (int i = 1; i <= m;i++){
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
e[++cnt] = Edge{x1 - 1, y1 - 1, 1, i};
e[++cnt] = Edge{x1 - 1, y2, 2, i};
e[++cnt] = Edge{x1 - 1, n, 3, i};
e[++cnt] = Edge{x2, y1 - 1, 4, i};
e[++cnt] = Edge{x2, y2, 5, i};
e[++cnt] = Edge{x2, n, 6, i};
e[++cnt] = Edge{n, y1 - 1, 7, i};
e[++cnt] = Edge{n, y2, 8, i};
e[++cnt] = Edge{n, n, 9, i};
}
sort(node + 1, node + 1 + n), sort(e + 1, e + cnt + 1);
for (int i = 1, j = 1; i <= cnt; i++){
while(j<=n && node[j].y<=e[i].y)
add(node[j++].x);
a[e[i].id][e[i].pos] = ask(e[i].x);
}
for (int i = 1; i <= m;i++){
tot = 0;
high[1] = a[i][1];
high[2] = a[i][2] - a[i][1];
high[3] = a[i][3] - a[i][2];
high[4] = a[i][4] - a[i][1];
high[5] = a[i][5] - a[i][4] - a[i][2] + a[i][1];
high[6] = a[i][6] - a[i][5] - a[i][3] + a[i][2];
high[7] = a[i][7] - a[i][4];
high[8] = a[i][8] - a[i][7] - a[i][5] + a[i][4];
high[9] = a[i][9] - a[i][8] - a[i][6] + a[i][5];
tot += high[5] * (n-high[5]);
tot += high[5] * (high[5] - 1) / 2;
tot += high[1] * (high[6] + high[8] + high[9]);
tot += high[2] * (high[4] + high[6] + high[7] + high[8] + high[9]);
tot += high[3] * (high[4] + high[7] + high[8]);
tot += high[4] * (high[6] + high[8] + high[9]);
tot += high[6] * (high[8] + high[7]);
cout << tot << endl;
}
}