团 大连网赛 1007 Friends and Enemies

//大连网赛 1007 Friends and Enemies
// 思路：思路很棒！
// 转化成最大二分图
// 团：点集的子集是个完全图
// 那么朋友圈可以考虑成一个团，原题就转化成用团去覆盖怎样最多。团至少是2个点，所以就是n*n/4

#include <bits/stdc++.h>
using namespace std;
#define LL long long
typedef pair<int,int> pii;
const double inf = 123456789012345.0;
const LL MOD =100000000LL;
const int N =1e4+10;
#define clc(a,b) memset(a,b,sizeof(a))
const double eps = 1e-7;
void fre() {freopen("in.txt","r",stdin);}
void freout() {freopen("out.txt","w",stdout);}
inline int read() {int x=0,f=1;char ch=getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch=getchar();}while(ch>='0'&&ch<='9') {x=x*10+ch-'0';ch=getchar();}return x*f;}

int main(){
    int n,m;
    while(~scanf("%d%d",&n,&m)){
       if(m>=(LL)n*n/4) puts("T");
       else puts("F");
    }
    return 0;
}