You are given a string S of length N consisting of lowercase English letters.

Process Q queries of the following two types:

* Type 1: change the i_q-th character of S to c_q. (Do nothing if the i_q-th character is already c_q.)
* Type 2: answer the number of different characters occurring in the substring of S between the l_q-th and r_q-th characters (inclusive).

Constraints

* N, Q, i_q, l_q, and r_q are integers.
* S is a string consisting of lowercase English letters.
* c_q is a lowercase English letter.
* 1 \leq N \leq 500000
* 1 \leq Q \leq 20000
* |S| = N
* 1 \leq i_q \leq N
* 1 \leq l_q \leq r_q \leq N
* There is at least one query of type 2 in each testcase.

Input

Input is given from Standard Input in the following format:


N
S
Q
Query_1
\vdots
Query_Q


Here, Query_i in the 4-th through (Q+3)-th lines is one of the following:


1 i_q c_q



2 l_q r_q


Output

For each query of type 2, print a line containing the answer.

Example

Input

7
abcdbbd
6
2 3 6
1 5 z
2 1 1
1 4 a
1 7 d
2 1 7


Output

3
1
5
#include <set>
#include <cstdio>
using namespace std;

const int N=5000005;

int n,Q;
char s[N];
set<int> Set[30];//Set[i]集合中记录第i个字母出现的位置 
set<int>::iterator it;

void init()
{scanf("%d",&n);
 scanf("%s",s+1);
 for (int i=0;i<26;i++)
   Set[i].insert(n+1);
 for (int i=1; i<=n; i++)
   Set[s[i]-'a'].insert(i); 
}

void work()
{int type,i,l,r,ans;
 char c;
 scanf("%d",&Q);
 for (int j=1;j<=Q;j++)
   {scanf("%d",&type);
    if (type==1)//替换操作 
      {scanf("%d %c",&i,&c);
       Set[s[i]-'a'].erase(i); 
	   Set[c-'a'].insert(i); 
	   s[i]=c;
	  }
	else//查询操作 
	  {scanf("%d %d",&l,&r);
	   ans=0;
	   for (int k=0;k<26;k++) 
	     {it=Set[k].lower_bound(l); 
		  if (*it<=r) ans++; 
		 }
	   printf("%d\n",ans);
	  }
   }
}
 
int main()
{
 init();
 work();
 return 0;
}
