~ phonism

## Tour de Byteotia.cpp
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

const int maxn = 1001000;
const int maxm = 2002000;

## Vouchers.cpp
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;

const int maxn = 1000011;

int n, m, x, maxm, a[maxn], pos[maxn];

## sgu499.cpp
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;

const int maxn = 1000010;

int n, maxx, a[100100], pos[maxn];

## hdu2426.cpp
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;

#define maxn 2110
#define maxm 2000001

## poj2195.cpp
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <string>
#include <queue>
using namespace std;

#define maxn 2110

## CF249E.cpp
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <string>
using namespace std;

const int maxn = 100100;

## CF296B.cpp
/*
a[i],b[i],c[i],d[i]分别表示 、ch1[i]<=ch2[i]的情况数、
ch1[i]>=ch2[i]的情况数、ch1[i]==ch2[i]的情况数看，
所有的情况数。那么根据容斥原理，
有ans = sum(d) - sum(a) - sum(b) + sum(c)
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

## CF3D.cpp
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;

priority_queue<pair<long long, long long> > q;
string str;

## POJ3468.cpp
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

const int maxn = 100100;

char op[10];

## CF341D.cpp
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

const int maxn = 1031;

int n, m, op, x1, y1, x2, y2;
	#include <cstdio>
	#include <cstring>
	#include <iostream>
	#include <algorithm>
	#include <vector>
	using namespace std;

	const int maxn = 1001000;
	const int maxm = 2002000;
	/*
	a[i],b[i],c[i],d[i]分别表示、ch1[i]<=ch2[i]的情况数、
	ch1[i]>=ch2[i]的情况数、ch1[i]==ch2[i]的情况数看，
	所有的情况数。那么根据容斥原理，
	有ans = sum(d) - sum(a) - sum(b) + sum(c)
	*/
	#include <cstdio>
	#include <cstring>
	#include <iostream>
	#include <algorithm>