Vova is again playing some computer game, now an RPG. In the game Vova's character received a quest: to slay the fearsome monster called Modcrab.

After two hours of playing the game Vova has tracked the monster and analyzed its tactics. The Modcrab has h2 health points and an attack power of a2. Knowing that, Vova has decided to buy a lot of strong healing potions and to prepare for battle.

Vova's character has h1 health points and an attack power of a1. Also he has a large supply of healing potions, each of which increases his current amount of health points by c1 when Vova drinks a potion. All potions are identical to each other. It is guaranteed that c1 > a2.

The battle consists of multiple phases. In the beginning of each phase, Vova can either attack the monster (thus reducing its health by a1) or drink a healing potion (it increases Vova's health by c1; Vova's health can exceed h1). Then, if the battle is not over yet, the Modcrab attacks Vova, reducing his health by a2. The battle ends when Vova's (or Modcrab's) health drops to 0 or lower. It is possible that the battle ends in a middle of a phase after Vova's attack.

Of course, Vova wants to win the fight. But also he wants to do it as fast as possible. So he wants to make up a strategy that will allow him to win the fight after the minimum possible number of phases.

Help Vova to make up a strategy! You may assume that Vova never runs out of healing potions, and that he can always win.

Input

The first line contains three integers h1, a1, c1 (1 ≤ h1, a1 ≤ 100, 2 ≤ c1 ≤ 100) — Vova's health, Vova's attack power and the healing power of a potion.

The second line contains two integers h2, a2 (1 ≤ h2 ≤ 100, 1 ≤ a2 < c1) — the Modcrab's health and his attack power.

Output

In the first line print one integer n denoting the minimum number of phases required to win the battle.

Then print n lines. i-th line must be equal to HEAL if Vova drinks a potion in i-th phase, or STRIKE if he attacks the Modcrab.

The strategy must be valid: Vova's character must not be defeated before slaying the Modcrab, and the monster's health must be 0 or lower after Vova's last action.

If there are multiple optimal solutions, print any of them.

Examples

Input

10 6 100
17 5


Output

4
STRIKE
HEAL
STRIKE
STRIKE


Input

11 6 100
12 5


Output

2
STRIKE
STRIKE

Note

In the first example Vova's character must heal before or after his first attack. Otherwise his health will drop to zero in 2 phases while he needs 3 strikes to win.

In the second example no healing needed, two strikes are enough to get monster to zero health and win with 6 health left.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.BitSet;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;

import javax.management.Query;
import javax.swing.plaf.nimbus.NimbusLookAndFeel;

import java.util.PriorityQueue;
import java.util.Set;
import java.util.Stack;
import java.util.StringTokenizer;
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.Double.parseDouble;

public class Solution { 
 	static  int mod=1000000007;

	public static void main(String[] args) throws IOException, InterruptedException {
		PrintWriter out=new PrintWriter(System.out); 
		BufferedReader br=new  BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st;
	    //reviser complement 2
		//hexadecimal
	    st=new StringTokenizer(br.readLine());
	    int h1=parseInt(st.nextToken());
	    int a1=parseInt(st.nextToken());
	    int c1=parseInt(st.nextToken());
	    st=new StringTokenizer(br.readLine());
	    int h2=parseInt(st.nextToken());
	    int a2=parseInt(st.nextToken());
	    StringBuilder sb=new StringBuilder();
	    int count=0;
	    while(h2>0){
	    	++count;
	    	if(h2 - a1 <= 0){
	    		sb.append("STRIKE\n");
	    		break;
	    	}
	    	else if(h1 - a2 <= 0){
	    		sb.append("HEAL\n");
	    		h1+=c1;
	    	}
	    	else{
	    		h2 -= a1;
	    		sb.append("STRIKE\n");
	    	}
	    	h1-=a2;
	    }
	    sb.deleteCharAt(sb.length()-1);
	    out.println(count);
	    out.println(sb);
	    
			   
		
		
		
		out.close();
		
	
	
	
	
	}
	static boolean answer(int n){
		if(n < 0) return false;
		else if(n == 0) return true;
		return (answer(n - 3) || answer(n - 7));
		
	}
	static int[][] createMatrix(int n, int m){
		int[][] matrix=new int[n][m];
		for(int i=0 ; i < n   ;++i){
			for(int k = 0; k < m;++k){
				if(k==0 || k==1) matrix[i][k]=i*m + k+1;
				else if (i%2 == 0){
					int inter=matrix[i][k-2];
					matrix[i][k] = inter;
					matrix[i][k-2]=i*m + k+1;
				}else{
					matrix[i][k]=i*m + k+1;
				}
			}
		}
		
		
		for(int i = 1; i < n; i+=2){
			for(int k=0; k < m; k+=2){
				int inter=matrix[i][k];
				matrix[i][k]=matrix[i-1][k];
				matrix[i-1][k]=inter;
			}
		}
		return matrix;
	}
	static boolean nim_game(int n){
		int[] win_pos=new int[]{2, 3, 5, 4};
		if(n > 0){
			for(int x: win_pos){
				if(x == n) return true;
			}
			if(n == 1) return false;
			if(!nim_game(n - 5)|| !nim_game(n - 3)|| !nim_game(n - 2)) return true;
		}
		return false;
	}
	static int xor_range(int n){
		if(n == 0) return 0;
		int r= n% 4;
		if(r == 0) return n; 
		if(r == 1) return 1;
		if(r == 2) return n+1;
		return 0;
	}
	static void sol(int[] arr, int index, int time, int k, HashMap<Integer, Integer> map, int sum){
		if( time == k+1){ 
				if(map.containsKey(sum)) map.put(sum, map.get(sum) + 1);
				else map.put(sum, 1);
		}
		
		if(time <= k){
			for(int i=index ; i  < arr.length; ++i){
				sol(arr, i + 1 , time + 1 , k, map, sum + arr[i]);
			}
		}
	}
	
	static int combinaison(double n, double k){//c(n, k)= (k+1)/(n-k)
		double res= 1;
		for(int i= 1; i<=k; ++i){
			res *= (n-i+1)/(i);
		}
		return (int)res;
	}
	static int countmin(int a, int b, int c){
		//Given a sequence of three binary sequences A, B and C of N bits. Count the minimum bits
		//required to flip in A and B such that XOR of A and B is equal to C
		int res= a ^ b ^ c;
		int count=0;
		while(res != 0){
			++count;
			res&=res - 1;
		}
		return count;
	}
	static int countval(int n){//Given a positive integer n, find count of positive integers i such that 0 <= i <= n and n+i = n^i
		int count=0;
		while(n != 0){
			if((n & 1) == 0) ++count;
		}
		return 1<<count;
	}
	static int firstright(int x){
		
		int y=1;
		int pos=1;
		while((x&y) == 0){
			pos++;
			y<<=1;
		}
		return pos;
	}
	
	static int atob(int a, int b){
		int c= a^b;
		int count=0;
		while(c != 0){
			c&=(c-1);
			++count;
		}
		return count;
	}
	  
	
	static int setBits(int x){
		int count=0;
		if(x == 0) return 0;
		while(x != 0){
			if(x % 2 == 1){
				--x;
				++count;
			}
			x/=2;
			
		}
		return count;
	}
	static boolean isPower4(long x){
			if((x & (x-1)) != 0) return false;
			int count=0;
			while(x != 1){
				x>>=1;
				++count;
			}
			
			if(count % 2 == 0) return true;
			return false;
	
	}
	
	
	
	static long addOne(long x){// add one using bit manipulation
	    long y = 1; 
		while((x & y) != 0){
			x ^= y;
			y<<=1;
		}
		
		return x ^ y;
	}
	
	static int compute_xor_range(int n){// compute xor from 1 to n
		int r= n % 4;
		if(r== 0) return n;
		else if(r== 1) return 1;
		else if(r== 2) return n+1;
		else return 0;
				
	}
	
	static long number_combinaisons(long n){//n cardinal of set 
		//fonction return number of combinaisons
		return (1 << n) ;
	}
	
	static boolean isPoweroftwo(long x){
		//check a number is power of two 
		//we use x & (x-1) to eliminate the first set bit
		
		return (x!=0 && (x & (x-1)) == 0)?true:false;
	}
	
	static long findxor(long[] Set,long n){// this function find xor of xor's of all subsets
		//n is length of set array
		if(n >1 ) return 0;
		else return Set[0];
	}
	// pour generer un nombre entier a partir de binaire int a=0b111;
	
	static long MSB(long n){// most significant bit
		while((n&(n-1)) != 0){
			n &= n-1;
		}
		return n;
	}
	
	static long LSB(long n){
	      if(n == 0) return 0;
	      return n-(n & (n-1));
	}
	
	
	 
	
		
}




