template.cpp

#include <iostream>
#include <algorithm>
#include <random>
#include <functional>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <chrono>
#include <iomanip>

// #pragma O(2)
// #define int long long
// 返回x的二进制中1的个数
#define ppc(x) __builtin_popcount(x)
// 返回x的二进制中前导0的个数
#define clz(x) __builtin_clz(x)
// 返回x的二进制中后导0的个数
#define ctz(x) __builtin_ctz(x)
// lg(8) = 3, lg(16) = 4, 计算x的以2为底的对数的向下取整, 即x的最高位1的位置
#define lg(x) __lg(x)

using namespace __gnu_pbds;
using namespace std;

using LL = long long;
using PII = pair<int, int>;
using PLL = pair<LL, LL>;
using PDI = pair<double, int>;
using PDD = pair<double, double>;
using ULL = unsigned long long;
template<class T> using sset = tree<T, null_type, less<>, rb_tree_tag, tree_order_statistics_node_update>;

const int mod = 998244353;
const int N = 1e5 + 10, M = 2e5 + 10;
const int INF = 0x3f3f3f3f;
const LL LLF = 0x3f3f3f3f3f3f3f3f;

// 用于检测程序占用内存大小
bool Mbe;

// 随机数生成器
// mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rnd(1064);
int rd(int l, int r) {return rnd() % (r - l + 1) + l;}
double rdd(double l, double r) {return rnd() * (r - l) / rnd.max() + l;}

// 取模运算
void plut(int &x, int y) {x += y, x >= mod && (x -= mod);}
void subt(int &x, int y) {x -= y, x < 0 && (x += mod);}
void mult(int &x, int y) {x = (LL)x * y % mod;}
int plu(int x, int y) {return x + y >= mod ? x + y - mod : x + y;}
int sub(int x, int y) {return x - y < 0 ? x - y + mod : x - y;}
int mul(int x, int y) {return (LL)x * y % mod;}

// 判断大小
template<typename T, typename U>
void chkmin(T &a, U b) {if (a > b) a = b;}
template<typename T, typename U>
void chkmax(T &a, U b) {if (a < b) a = b;}

// 快速幂
int qpow(int a, int b) {
    int res = 1;
    while (b) {
        if (b & 1) mult(res, a);
        mult(a, a), b >>= 1;
    }
    return res;
}

// 最大公约数
int gcd(int a, int b) {return b ? gcd(b, a % b) : a;}

// 逆元
int inv(int x) {return qpow(x, mod - 2);}

// 二分
int bsearch(int l, int r, function<bool(int)> f) {
    while (l < r) {
        int mid = (l + r) >> 1;
        if (f(mid)) r = mid;
        else l = mid + 1;
    }
    return l;
}

void setIO() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
}

// 读入优化
template<typename T>
inline T read(){
	T a = 0;
    bool s = 0;
	char ch = getchar();
	while (ch > '9' || ch < '0'){
		if (ch == '-') s ^= 1;
		ch = getchar();
	}
	while (ch >= '0' && ch <= '9'){
		a = (a << 3) + (a << 1) + (ch ^ 48);
		ch = getchar();
	}
	return s ? -a : a;
}

// 输出优化
inline void pf(LL x) {
    if (x < 0) putchar('-'), x = -x;
    if (x > 9) pf(x / 10);
    putchar(x % 10 + 48);
}

/*
// 求组合数
const int Z = 1e6 + 10;
int fc[Z], ifc[Z];
int bin(int n, int m) {
    if (n < m) return 0;
    return 1ll * fc[n] * ifc[m] % mod * ifc[n - m] % mod;
}
void init_fac(int Z) {
    for (int i = fc[0] = 1; i < Z; i++) fc[i] = 1ll * fc[i - 1] * i % mod;
    ifc[Z - 1] = qpow(fc[Z - 1], mod - 2);
    for (int i = Z - 2; ~i; i--) ifc[i] = 1ll * ifc[i + 1] * (i + 1) % mod;
}
*/

void solve() {

}

bool Med;
int main() {
    fprintf(stderr, "%.3lf MB\n", (&Med - &Mbe) / 1048576.0);
    // setIO();
    int T = 1;
    cin >> T;
    while (T --) solve();
    cerr << 1e3 * clock() / CLOCKS_PER_SEC << " ms\n";
    return 0;
}