In computability theory, the halting problem is the problem of determining, from a description of an arbitrary computer program, whether the program will finish running (i.e., halt) or continue to run forever.

Alan Turing proved in 1936 that a general algorithm to solve the halting problem cannot exist, but DreamGrid, our beloved algorithm scientist, declares that he has just found a solution to the halting problem in a specific programming language – the Dream Language!

Dream Language is a programming language consists of only 5 types of instructions. All these instructions will read from or write to a 8-bit register r, whose value is initially set to 0. We now present the 5 types of instructions in the following table. Note that we denote the current instruction as the i-th instruction.

Instruction Description add v Add v to the register r. As r is a 8-bit register, this instruction actually calculates (r+v)mod256 and stores the result into r, i.e. r←(r+v)mod256. After that, go on to the (i+1)-th instruction. beq v k If the value of r is equal to v, jump to the k-th instruction, otherwise go on to the (i+1)-th instruction. bne v k If the value of r isn’t equal to v, jump to the k-th instruction, otherwise go on to the (i+1)-th instruction. blt v k If the value of r is strictly smaller than v, jump to the k-th instruction, otherwise go on to the (i+1)-th instruction. bgt v k If the value of r is strictly larger than v, jump to the k-th instruction, otherwise go on to the (i+1)-th instruction. A Dream Language program consisting of n instructions will always start executing from the 1st instruction, and will only halt (that is to say, stop executing) when the program tries to go on to the (n+1)-th instruction.

As DreamGrid’s assistant, in order to help him win the Turing Award, you are asked to write a program to determine whether a given Dream Language program will eventually halt or not.

Input

There are multiple test cases. The first line of the input is an integer T, indicating the number of test cases. For each test case:

The first line contains an integer n (1≤n≤10 ​4 ​​ ), indicating the number of instructions in the following Dream Language program.

For the following n lines, the i-th line first contains a string s (s∈{“add”,“beq”,“bne”,“blt”,“bgt”}), indicating the type of the i-th instruction of the program.

If s equals to “add”, an integer v follows (0≤v≤255), indicating the value added to the register; Otherwise, two integers v and k follow (0≤v≤255, 1≤k≤n), indicating the condition value and the destination of the jump. It’s guaranteed that the sum of n of all test cases will not exceed 10 ​5 ​​ .

Output

For each test case output one line. If the program will eventually halt, output “Yes” (without quotes); If the program will continue to run forever, output “No” (without quotes).

Sample Input

4 2 add 1 blt 5 1 3 add 252 add 1 bgt 252 2 2 add 2 bne 7 1 3 add 1 bne 252 1 beq 252 1 Sample Output

Yes Yes No No

Hint

For the second sample test case, note that r is a 8-bit register, so after four “add 1” instructions the value of r will change from 252 to 0, and the program will halt.

For the third sample test case, it’s easy to discover that the value of r will always be even, so it’s impossible for the value of r to be equal to 7, and the program will run forever.

題意：

有一種演算法流程還是什麼東西，有五種操作，問這種演算法最後是否能形成死迴圈，如果可以輸出No，否則Yes。

思路：

直接模擬即可，判斷死迴圈我用的是判斷當前步驟執行的次數是否大於256。

#include <iostream>
#include <string>
#include <map>
#include <vector>
#include <set>
#include <list>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <queue>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 1e4 + 10;
const int maxx = 1e4 + 10;
const int mod = 1e9 + 7;
const double PI = acos(-1);
const double eps = 1e-8;
int T, n, cnt[maxn];
struct instruction {
    string op;
    int v, k;
} ins[maxn];
bool solve(int p) {
    queue < pair <int, int> > q;
    q.push(make_pair(p, 0));
    while (!q.empty()) {
        pair <int, int> now = q.front(), next;
        q.pop();
        cnt[now.first]++;
        if (cnt[now.first] > 256) return false;
        if (now.first > n) return true;
        if (ins[now.first].op == "add") {
            next.second = (now.second + ins[now.first].v) % 256;
            next.first = now.first + 1;
            q.push(next);
        }
        else if (ins[now.first].op == "beq") {
            if (now.second == ins[now.first].v) next.first = ins[now.first].k;
            else next.first = now.first + 1;
            next.second = now.second;
            q.push(next);
        }
        else if (ins[now.first].op == "bne") {
            if (now.second != ins[now.first].v) next.first = ins[now.first].k;
            else next.first = now.first + 1;
            next.second = now.second;
            q.push(next);
        }
        else if (ins[now.first].op == "blt") {
            if (now.second < ins[now.first].v) next.first = ins[now.first].k;
            else next.first = now.first + 1;
            next.second = now.second;
            q.push(next);
        }
        else {
            if (now.second > ins[now.first].v) next.first = ins[now.first].k;
            else next.first = now.first + 1;
            next.second = now.second;
            q.push(next);
        }
    }
    return false;
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> T;
    while (T--) {
        memset(cnt, 0, sizeof cnt);
        cin >> n;
        for (int i = 1; i <= n; i++) {
            cin >> ins[i].op;
            if (ins[i].op == "add") cin >> ins[i].v;
            else cin >> ins[i].v >> ins[i].k;
        }
        if (solve(1)) cout << "Yes\n";
        else cout << "No\n";
    }
}