cin>>n>>k;
ll cur=1;
k--;
while(k){
	ll steps=0, first=cur, last=cur;
	while(first<=n){ //steps表示前缀为cur，有多少个数字<=n
		steps+=min(last,n)-first+1;
		first=first*10;
		last=last*10+9;
	}
	if(steps<=k){
		k-=steps;
		cur++;
	}else{
		cur=cur*10;
		k--;
	}
}
cout<<cur;