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;
#define All(v) (v).begin(),(v).end()

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,q;
    string s;
    cin>>n>>s;
    set<int> st[26];
    for(int i=0;i<n;i++)st[s[i]-'a'].insert(i+1);
    cin>>q;
    for(int i=0;i<q;i++){
        int t;
        cin>>t;
        if(t==1){
            int idx;
            char c;
            cin>>idx>>c;
            st[s[idx-1]-'a'].erase(idx);
            st[c-'a'].insert(idx);
            s[idx-1]=c;
        }else{
            int res=0;
            int l,r;
            cin>>l>>r;
            for(int j=0;j<26;j++){
                auto it=st[j].lower_bound(l);
                if(it!=st[j].end()&&*it<=r)res++;
            }
            cout<<res<<endl;
        }
    }
}