# 右端の山に石が2個以上あれば先手が勝つ
# 右端なしの局面で先手必勝なら先手は全石を取る、後手必勝なら先手は1個残しとするから
# 右端の山に石が1個ならば、右端なしの局面で後手先手と考えればいい
# 結局右端をkとして、2個以上ある山のindexを見て
# k-indexが偶数なら先手勝ち、奇数なら後手勝ち
# セグメントツリーで高速化

# ACL for Python 
# https://github.com/shakayami/ACL-for-python/blob/master/segtree.py
# https://github.com/shakayami/ACL-for-python
# https://github.com/shakayami/ACL-for-python/wiki/segtree
# セグメントツリー：リストの要約値(たとえばMAX)を記録
 
class segtree():
    n=1
    size=1
    log=2
    d=[0]
    op=None
    e=10**15
    def __init__(self,V,OP,E):
        self.n=len(V)
        self.op=OP
        self.e=E
        self.log=(self.n-1).bit_length()
        self.size=1<<self.log
        self.d=[E for i in range(2*self.size)]
        for i in range(self.n):
            self.d[self.size+i]=V[i]
        for i in range(self.size-1,0,-1):
            self.update(i)
    def set(self,p,x):
        assert 0<=p and p<self.n
        p+=self.size
        self.d[p]=x
        for i in range(1,self.log+1):
            self.update(p>>i)
    def get(self,p):
        assert 0<=p and p<self.n
        return self.d[p+self.size]
    def prod(self,l,r):
        assert 0<=l and l<=r and r<=self.n
        sml=self.e
        smr=self.e
        l+=self.size
        r+=self.size
        while(l<r):
            if (l&1):
                sml=self.op(sml,self.d[l])
                l+=1
            if (r&1):
                smr=self.op(self.d[r-1],smr)
                r-=1
            l>>=1
            r>>=1
        return self.op(sml,smr)
    def all_prod(self):
        return self.d[1]
    def max_right(self,l,f):
        assert 0<=l and l<=self.n
        assert f(self.e)
        if l==self.n:
            return self.n
        l+=self.size
        sm=self.e
        while(1):
            while(l%2==0):
                l>>=1
            if not(f(self.op(sm,self.d[l]))):
                while(l<self.size):
                    l=2*l
                    if f(self.op(sm,self.d[l])):
                        sm=self.op(sm,self.d[l])
                        l+=1
                return l-self.size
            sm=self.op(sm,self.d[l])
            l+=1
            if (l&-l)==l:
                break
        return self.n
    def min_left(self,r,f):
        assert 0<=r and r<self.n
        assert f(self.e)
        if r==0:
            return 0
        r+=self.size
        sm=self.e
        while(1):
            r-=1
            while(r>1 & (r%2)):
                r>>=1
            if not(f(self.op(self.d[r],sm))):
                while(r<self.size):
                    r=(2*r+1)
                    if f(self.op(self.d[r],sm)):
                        sm=self.op(self.d[r],sm)
                        r-=1
                return r+1-self.size
            sm=self.op(self.d[r],sm)
            if (r& -r)==r:
                break
        return 0
    def update(self,k):
        self.d[k]=self.op(self.d[2*k],self.d[2*k+1])
    def __str__(self):
        return str([self.get(i) for i in range(self.n)])

def segfunc(x, y):
    return max(x, y)    

N, Q = map(int, input().split())
A = list(map(int, input().split()))

B = []
for i in range(N):
    if A[i] != 1:
        B.append(i)
    else:
        B.append(-1)

ST = segtree(B, segfunc, 0)
        
ans = []

for q in range(Q):
    t, x, y = map(int, input().split())
    if t == 1:
        if y == 1:
            ST.set(x-1, -1)
        else:
            ST.set(x-1, x-1)
    else:
        a = ST.prod(x-1, y)
        if a == -1:
            if (y-x)%2 == 0:
                ans.append('F')
            else:
                ans.append('S')
        else:
            if (y-1-a)%2 == 0:
                ans.append('F')
            else:
                ans.append('S')

for a in ans:
    print(a)