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;

int main()
{
    int N;
    cin>>N;
    string s;
    cin>>s;
    int n=s.size();
    vector<set<int>>v(26);
    for(int i=0;i<n;++i)
    {
        v[s[i]-'a'].insert(i);
    }
    int q;
    cin>>q;
    while(q--){

        int a;
        cin>>a;

        if(a==1){

            int pos; char ch;
            cin>>pos>>ch;
            --pos;
            v[s[pos]-'a'].erase(pos);
            s[pos]=ch;
            v[s[pos]-'a'].insert(pos);
        }
        else{

            int l,r;
            cin>>l>>r;
            --l,--r;
            int ans=0;
            for(int i=0;i<26;++i){

                auto it=v[i].lower_bound(l);
                if(it!=v[i].end()&& *it<=r) ++ans;
            }
           cout<<ans<<endl;
    }
 }
}