数据结构（线段树）：CodeForces 145E Lucky Queries

E. Lucky Queries

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Petya brought home string s with the length of n. The string only consists of lucky digits. The digits are numbered from the left to the right starting with 1. Now Petya should execute m queries of the following form:

• switch l r — "switch" digits (i.e. replace them with their opposites) at all positions with indexes from l to r, inclusive: each digit 4 is replaced with 7 and each digit 7 is replaced with 4 (1 ≤ l ≤ r ≤ n);
• count — find and print on the screen the length of the longest non-decreasing subsequence of string s.

Subsequence of a string s is a string that can be obtained from s by removing zero or more of its elements. A string is called non-decreasing if each successive digit is not less than the previous one.

Help Petya process the requests.

Input

The first line contains two integers n and m (1 ≤ n ≤ 10⁶, 1 ≤ m ≤ 3·10⁵) — the length of the string s and the number of queries correspondingly. The second line contains n lucky digits without spaces — Petya's initial string. Next m lines contain queries in the form described in the statement.

Output

For each query count print an answer on a single line.

Examples

Input

2 3
47
count
switch 1 2
count

Output

2
1

Input

3 5
747
count
switch 1 1
count
switch 1 3
count

Output

2
3
2

Note

In the first sample the chronology of string s after some operations are fulfilled is as follows (the sought maximum subsequence is marked with bold):

1. 47
2. 74
3. 74

In the second sample:

1. 747
2. 447
3. 447
4. 774
5. 774

　　比较好写……

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 using namespace std;
 const int maxn=;
 char s[maxn];
 int num[maxn],Mark[maxn<<];
 int M1[maxn<<],M2[maxn<<];
 int M3[maxn<<],M4[maxn<<];
 int tot[maxn<<],n,Q;
 //M1:00 M2:11 M3:01 M4:10 

 void Swich(int x){
     swap(M1[x],M2[x]);
     swap(M3[x],M4[x]);
     Mark[x]^=;
 }

 void Push_down(int x,int l,int r){
     if(!Mark[x]||l==r)return;
     Swich(x<<);Swich(x<<|);
     Mark[x]=;
 }

 void Push_up(int x){
     M1[x]=M1[x<<]+M1[x<<|];
     M2[x]=M2[x<<]+M2[x<<|];
     M3[x]=max(M1[x<<]+M2[x<<|],max(M1[x<<]+M3[x<<|],M3[x<<]+M2[x<<|]));
     M4[x]=max(M2[x<<]+M1[x<<|],max(M4[x<<]+M1[x<<|],M2[x<<]+M4[x<<|]));
 }

 void Build(int x,int l,int r){
     if(l==r){
         M1[x]=num[l]^;
         M2[x]=num[l];
         return;
     }
     int mid=(l+r)>>;
     Build(x<<,l,mid);
     Build(x<<|,mid+,r);
     Push_up(x);
 }

 void Update(int x,int l,int r,int a,int b){
     Push_down(x,l,r);
     if(l>=a&&r<=b){
         Swich(x);
         return;
     }
     int mid=(l+r)>>;
     if(mid>=a)Update(x<<,l,mid,a,b);
     if(mid<b)Update(x<<|,mid+,r,a,b);
     Push_up(x);
 }

 char op[];
 int main(){
     scanf("%d%d",&n,&Q);
     scanf("%s",s+);
     for(int i=;i<=n;i++){
         num[i]=s[i]==''?:;
     }
     Build(,,n);
     int a,b;
     while(Q--){
         scanf("%s",op);
         if(op[]=='s'){
             scanf("%d%d",&a,&b);
             Update(,,n,a,b);
         }
         else
             printf("%d\n",max(max(M1[],M2[]),M3[]));
     }
     return ;
 }