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++)
typedef long long ll;
using namespace std;
int main(void){
    int n,q;string s;cin>>n>>s>>q;
    vector<set<int>> is(26);
    rep(i,n) is[s[i]-'a'].insert(i);
    rep(qi,q){
        int ty;cin>>ty;
        if(ty==1){
            int i;char r;cin>>i>>r;i--;
            is[s[i]-'a'].erase(i);
            s[i]=r;
            is[s[i]-'a'].insert(i);
        }else{
            int l,r;cin>>l>>r;l--;r--;
            int res=0;
            rep(i,26){
                auto itr= is[i].lower_bound(l);
                if(itr!=is[i].end()&&*itr<=r)res++;
            }
            cout<<res<<endl;
        }
    }
}
