B-one box

Description

You have a box numbered 1, 2, 3,… from left to right ,..., N. You can execute four commands:

1 x y indicates moving box X to the left of box Y (ignore this command if X is already on the left of box Y ).
2 x y indicates moving box X to the right of box Y (ignore this command if X is already on the right of box Y ).
3 x y indicates the position of swap box X and Y.
4 indicates that the entire chain is reversed.

The command must be valid, that is, X is not equal to y. For example, when n is 6 and 1 1 4 is executed in the initial state, the box sequence is 2 3 1 4 5 6. Next, execute 2 3 5, and the box sequence is changed to 2 1 4 5 3 6. Then execute 3 1 6 and get 2 6 4 5 3 1. Finally execute 4 and get 1 3 5 4 6 2.

 

Input

The input contains no more than 10 groups of data. The number of first behavior boxes in each group is n and the number of command lines is m (1 <= n, m <= 100,000). The following M lines contain one command per line.

 

Output

Each data group outputs a row, that is, the sum of box numbers at all odd positions. Positions from left to right are numbered 1 ~ N.

 

Sample Input

6 41 1 42 3 53 1 646 31 1 42 3 53 1 6100000 14 

Sample output

Case 1: 12 Case 2: 9 case 3: 2500050000

The difficulty lies in the fourth type of operation, the fourth type of operation is the reversal, and the effect is the follow-up exchange. In fact, there is no need to perform the operation. You only need to record the number of reverse operations. When the number of reversal operations is odd, the operations with flag = 1 are converted to operations with flag = 2, and those with flag = 2 are converted to operations with flag = 1, while those with other operations remain unchanged.
Because after the reversal, the front of a number is changed to the back, and the back is changed to the front. There is also a ring to facilitate the start and end points.

1 # include "stdio. H "2 # include" string. H "3 const int MS = 100004; 4 int left [100004]; 5 Int right [100004]; 6 int * tmp1, * tmp2; 7 inline void Link (int, int B) 8 {9 right [a] = B; 10 left [B] = A; 11} 12 INT main () 13 {14 int n, m, I, J, k, flag, X, Y, P = 1, CNT, next; 15 while (scanf ("% d", & N, & M )! = EOF) 16 {17 for (I = 1; I <= N; I ++) 18 {19 left [I] = I-1; 20 right [I] = I + 1; 21} 22 right [N] = 0; // forms a ring to facilitate searching for the Start Node and End Node 23 CNT = 0; 24 while (M --) 25 {26 scanf ("% d", & flag); 27 if (flag! = 4) 28 scanf ("% d", & X, & Y); 29 else 30 CNT ++; 31 if (CNT & 1) & flag <3) 32 flag = 3-flag; 33 If (flag = 1 & (right [x]! = Y) 34 {35 if (right [y] = x) 36 {37 int T1 = left [X], T2 = right [X], t3 = left [Y], T4 = right [y]; 38 Link (T3, x); 39 Link (x, y); 40 Link (Y, T2 ); 41} 42 else 43 {44 Link (left [X], right [x]); 45 Link (left [Y], x); 46 Link (x, y ); 47} 48} 49 else if (flag = 2 & right [y]! = X) 50 {51 if (right [x] = y) 52 {53 int T1 = left [X], T2 = right [X], t3 = left [Y], T4 = right [y]; 54 Link (T1, Y); 55 Link (Y, x); 56 Link (x, T4 ); 57} 58 else 59 {60 Link (left [X], right [x]); 61 Link (x, right [y]); 62 Link (Y, X ); 63} 64} 65 else if (flag = 3) 66 {67 If (right [x] = y) 68 {69 int T1 = left [X], t2 = right [X], T3 = left [Y], T4 = right [y]; 70 Link (T1, Y); 71 Link (Y, X ); 72 Link (x, T4); 73} 74 else if (right [y] = x) 75 {76 int T1 = left [X], T2 = right [X], t3 = left [Y], T4 = right [y]; 77 Link (T3, x); 78 Link (x, y); 79 Link (Y, T2 ); 80} 81 else {82 int T1 = left [X], T2 = right [X], T3 = left [Y], T4 = right [y]; 83 Link (T1, y); 84 Link (Y, T2); 85 Link (T3, x); 86 Link (x, T4); 87} 88} 89} 90 long ans = 0; 91 If (CNT & 1) 92 {93 tmp2 = right; 94 tmp1 = left; 95} 96 else 97 {98 tmp2 = left; 99 tmp1 = right; 100} 101 for (I = 1; I <= N; I ++) 102 {103 If (tmp2 [I] = 0) 104 {105 for (k = 1; i; I = tmp1 [I], K ++) 106 If (K & 1) 107 ans + = I; 108 printf ("case % d: % LLD \ n ", P ++, ANS); 109 break; 110} 111} 112 return 0; 113}

 

B-one box