#include "rect.h"
#include <bits/stdc++.h>
using namespace std;
template<class T, class U>
void ckmin(T &a, U b)
{
if (a > b) a = b;
}
template<class T, class U>
void ckmax(T &a, U b)
{
if (a < b) a = b;
}
#define y1 qkx
#define y2 wsm
#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)
#define SZ(x) ((int) (x).size())
#define ALL(x) (x).begin(), (x).end()
#define INF 1000000007
#define MAXN 2513
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;
int N, M;
vector<vi> grid;
int rt[MAXN][MAXN], dn[MAXN][MAXN], lt[MAXN][MAXN], up[MAXN][MAXN];
int pref[MAXN][MAXN];
bool subtask6 = true;
vector<vector<vi> > col, row;
set<pii> edges;
ll ans;
int psum(int x1, int x2, int y1, int y2)
{
return pref[x2 + 1][y2 + 1] - pref[x1][y2 + 1] - pref[x2 + 1][y1] + pref[x1][y1];
}
ll count_rectangles(vector<vi> a)
{
N = SZ(a); M = SZ(a[0]); grid = a;
//BE SUPER CAREFUL ABOUT EQUALITY CASE.
//for each guy, find the next guy right of it that's greater th an it.
//ok fix the starting column #.
FOR(i, 0, N)
{
vpi cand; cand.PB({INF, M});
FORD(j, M, 0)
{
while(cand.back().fi < grid[i][j]) cand.pop_back();
rt[i][j] = cand.back().se;
cand.PB({grid[i][j], j});
}
cand.clear(); cand.PB({INF, -1});
FOR(j, 0, M)
{
while(cand.back().fi < grid[i][j]) cand.pop_back();
lt[i][j] = cand.back().se;
cand.PB({grid[i][j], j});
}
}
FOR(i, 0, M)
{
vpi cand; cand.PB({INF, N});
FORD(j, N, 0)
{
while(cand.back().fi < grid[j][i]) cand.pop_back();
dn[j][i] = cand.back().se;
cand.PB({grid[j][i], j});
}
cand.clear(); cand.PB({INF, -1});
FOR(j, 0, N)
{
while(cand.back().fi < grid[j][i]) cand.pop_back();
up[j][i] = cand.back().se;
cand.PB({grid[j][i], j});
}
}
// FOR(i, 0, N)
// {
// FOR(j, 0, M)
// {
// cerr << up[i][j] << ' ';
// }
// cerr << endl;
// }
FOR(i, 0, N - 1)
{
edges.clear();
FOR(j, 0, M)
{
if (lt[i + 1][j] != -1 && lt[i + 1][j] < j - 1) edges.insert({lt[i + 1][j], j});
if (rt[i + 1][j] != M && rt[i + 1][j] > j + 1) edges.insert({j, rt[i + 1][j]});
}
// cerr << "FROM ROW " << i << endl;
FOR(j, i + 2, N)
{
//find all edges which they share in common.
// cerr << "ROWS " << i << " -> " << j << endl;
// cerr << "EDGES\n";
// for (pii p : edges)
// {
// cerr << p.fi << ' ' << p.se << endl;
// }
vpi ks;
FOR(k, 0, M)
{
if (dn[i][k] == j || up[j][k] == i)
{
// cerr << "HI\n";
if (ks.empty() || ks.back().se != k - 1)
{
ks.PB({k, k});
}
else
{
ks.back().se++;
}
// cerr << "BYE\n";
}
}
// cerr << "KS\n";
// for (pii p : ks)
// {
// cerr << p.fi << ',' << p.se << endl;
// }
for (pii p : ks)
{
for (pii q : edges)
{
if (p.fi <= q.fi + 1 && q.se - 1 <= p.se)
{
// cerr << "YAY " << i << ' ' << j << ' ' << q.fi << ' ' << q.se << endl;
ans++;
}
}
}
for (auto it = edges.begin(); it != edges.end(); )
{
int x = it -> fi, y = it -> se;
if (rt[j][x] != y && lt[j][y] != x) it = edges.erase(it);
else it++;
}
//ok now find these ranges.
}
}
return ans;
}
