C. Replacement

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Daniel has a string s, consisting of lowercase English letters and period signs (characters '.'). Let's define the operation of replacementas the following sequence of steps: find a substring ".." (two consecutive periods) in string s, of all occurrences of the substring let's choose the first one, and replace this substring with string ".". In other words, during the replacement operation, the first two consecutive periods are replaced by one. If string s contains no two consecutive periods, then nothing happens.

Let's define f(s) as the minimum number of operations of replacement to perform, so that the string does not have any two consecutive periods left.

You need to process m queries, the i-th results in that the character at position xi (1 ≤ xi ≤ n) of string s is assigned value ci. After each operation you have to calculate and output the value of f(s).

Help Daniel to process all queries.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 300 000) the length of the string and the number of queries.

The second line contains string s, consisting of n lowercase English letters and period signs.

The following m lines contain the descriptions of queries. The i-th line contains integer xi and ci (1 ≤ xi ≤ n, ci — a lowercas English letter or a period sign), describing the query of assigning symbol ci to position xi.

Output

Print m numbers, one per line, the i-th of these numbers must be equal to the value of f(s) after performing the i-th assignment.

Sample test(s)

input

10 3.b..bz....1 h3 c9 f

output

431

input

4 4.cc.2 .3 .2 a1 a

output

1311

Note

Note to the first sample test (replaced periods are enclosed in square brackets).

The original string is ".b..bz....".

• after the first query f(hb..bz....) = 4    ("hb[..]bz...."  →  "hb.bz[..].."  →  "hb.bz[..]."  →  "hb.bz[..]"  → "hb.bz.")
• after the second query f(hbс.bz....) = 3    ("hbс.bz[..].."  →  "hbс.bz[..]."  →  "hbс.bz[..]"  →  "hbс.bz.")
• after the third query f(hbс.bz..f.) = 1    ("hbс.bz[..]f."  →  "hbс.bz.f.")

Note to the second sample test.

The original string is ".cc.".

• after the first query: f(..c.) = 1    ("[..]c."  →  ".c.")
• after the second query: f(....) = 3    ("[..].."  →  "[..]."  →  "[..]"  →  ".")
• after the third query: f(.a..) = 1    (".a[..]"  →  ".a.")
• after the fourth query: f(aa..) = 1    ("aa[..]"  →  "aa.")

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #define lc idx<<1#define rc idx<<1|1#define lson l,mid,lc#define rson mid+1,r,rc#define N 300010using namespace std;int n,m;char s[N];struct node {    bool ok;   ///整段是否为‘*’    bool ls,rs; ///左右端点是否为‘*’    int num;   } tree[N<<2];void push_up(int idx,int l,int r) {    tree[idx].ok=tree[lc].ok&&tree[rc].ok;    if(tree[idx].ok) {        tree[idx].num=r-l;        tree[idx].ls=tree[idx].rs=1;    } else {        tree[idx].num=tree[lc].num+tree[rc].num;        if(tree[lc].rs&&tree[rc].ls)            tree[idx].num++;        tree[idx].ls=tree[lc].ls;        tree[idx].rs=tree[rc].rs;    }}void build(int l,int r,int idx) {    if(l==r) {        tree[idx].num=0;        if(s[l]=='.') {            tree[idx].ls=tree[idx].rs=1;            tree[idx].ok=1;        } else {            tree[idx].ls=tree[idx].rs=0;            tree[idx].ok=0;        }        return;    }    int mid=(l+r)>>1;    build(lson);    build(rson);    push_up(idx,l,r);}void update(int l,int r,int idx,int pos) {    if(l==r) {        if(s[l]=='.') {            tree[idx].ls=tree[idx].rs=1;            tree[idx].ok=1;        } else {            tree[idx].ls=tree[idx].rs=0;            tree[idx].ok=0;        }        return ;    }    int mid=(l+r)>>1;    if(pos<=mid) {        update(lson,pos);    } else {        update(rson,pos);    }    push_up(idx,l,r);}int main() {    //freopen("test.in","r",stdin);    while(~scanf("%d%d",&n,&m)) {        scanf("%s",s+1);        build(1,n,1);        char c[2];        int pos;        while(m--) {            scanf("%d%s",&pos,c);            if(c[0]=='.'&&s[pos]=='.') {                printf("%d\n",tree[1].num);                continue;            }            if(c[0]!='.'&&s[pos]!='.') {                printf("%d\n",tree[1].num);                s[pos]=c[0];                continue;            }            s[pos]=c[0];            update(1,n,1,pos);            printf("%d\n",tree[1].num);        }    }    return 0;}