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>
#define rep(i,n) for (int i=0; i<(n); ++i)
using namespace std;
using ll=long long;

int main(){
    int n,q;
    string s;
    cin>>n>>s>>q;
    vector<set<int>>se(26);
    rep(i,n)se[s[i]-'a'].insert(i);
    rep(i,q){
        int a,b; cin>>a>>b;
        b--;
        if(a==1){
            char c; cin>>c;
            se[s[b]-'a'].erase(b);
            s[b]=c;
            se[s[b]-'a'].insert(b);
        }
        else{
            int c; cin>>c;
            c--;
            int ans=0;
            rep(j,26){
                auto it=se[j].lower_bound(b);
                if(*it<=c&&it!=se[j].end())ans++;
            }
            printf("%d\n",ans);
        }
    }
}