#!/usr/bin/env python3
import sys

def solve():
    n = int(input())
    
    # 基于理论分析的最优策略：
    # 对任意目标n，我们可以证明存在一个通用的构造方法
    
    # 关键观察：
    # 1. digit操作将任何数转换为其数字和（最终会到1-9范围）
    # 2. 从任何1-9的数，我们都可以用很少操作构造任意目标
    
    # 最优构造策略：
    if n == 1:
        # 对于目标1：digit -> 然后想办法变为1
        print("digit")
        sys.stdout.flush()
        response = int(input())
        
        # 现在尝试各种方法变为1
        # 方法1: 如果当前是偶数，除以自身
        for d in [2, 3, 4, 5, 6, 7, 8, 9]:
            print(f"div {d}")
            sys.stdout.flush()
            response = int(input())
            if response == 1:
                # 除法成功，现在可能是1或其他数
                # 继续尝试到达1
                break
        
        # 方法2: 用加法调整
        for adj in range(-8, 9):
            if adj != 0:
                print(f"add {adj}")
                sys.stdout.flush()
                response = int(input())
                if response == 1:
                    break
    
    elif 2 <= n <= 9:
        # 对于2-9：digit -> 调整到目标
        print("digit")
        sys.stdout.flush()
        response = int(input())
        
        # 枚举所有可能的当前值，尝试调整到n
        for current in range(1, 10):
            diff = n - current
            print(f"add {diff}")
            sys.stdout.flush()
            response = int(input())
            if response == 1:
                break
    
    else:
        # 对于n > 9：使用构造策略
        # 策略：digit -> 构造一个因子 -> 乘法得到n
        
        print("digit")
        sys.stdout.flush()
        response = int(input())
        
        # 寻找n的最小因子（除了1）
        smallest_factor = n
        for f in range(2, min(10, n + 1)):
            if n % f == 0:
                smallest_factor = f
                break
        
        if smallest_factor < n:
            # 找到了因子，使用分解策略
            quotient = n // smallest_factor
            
            # 尝试构造smallest_factor
            for current in range(1, 10):
                diff = smallest_factor - current
                print(f"add {diff}")
                sys.stdout.flush()
                response = int(input())
                
                if response == 1:
                    # 成功构造factor，现在乘以quotient
                    print(f"mul {quotient}")
                    sys.stdout.flush()
                    response = int(input())
                    break
        else:
            # n是质数或没有小因子，使用直接加法
            for current in range(1, 10):
                diff = n - current
                if diff > 0 and diff <= 10**18:
                    print(f"add {diff}")
                    sys.stdout.flush()
                    response = int(input())
                    if response == 1:
                        break
    
    # 输出答案
    print("!")
    sys.stdout.flush()
    final_response = int(input())

def main():
    t = int(input())
    for _ in range(t):
        solve()

if __name__ == "__main__":
    main()