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<bits/stdc++.h>
using namespace std;
const int manx=5e5+50;
char a[manx];
set<int>s[50];
int x,l,r,q,n;
char c;
int main()
{
    int n; scanf("%d",&n);
    scanf("%s%d",a+1,&q);
    for(int i=1;i<=n;i++)
        s[a[i]-'a'].insert(i);
    while(q--)
    {
        scanf("%d",&x);
        if(x==2){
            scanf("%d%d",&l,&r);
            int ans=0;
            for(int i=0;i<26;i++)
            {
                auto it=s[i].lower_bound(l);
                if(it!=s[i].end()&&*it<=r) ans++;
            }
            printf("%d\n",ans);
        }
        else if(x==1){
            cin>>l>>c;
            s[a[l]-'a'].erase(l);
            s[c-'a'].insert(l);
            a[l]=c;
        }
    }
    return 0;
}
