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>
typedef long long ll;
using namespace std;
int main(){
    cin.tie(NULL); cout.tie(NULL);
    ios_base:: sync_with_stdio(false);
    string s;
    ll n,q;
    cin>>n>>s>>q;
    s = '$' + s;
    set<ll> st[26];
    for(ll i=1;i<=n;i++)st[s[i]-'a'].insert(i);
    while(q--){
        ll type;cin>>type;
        if(type==1){
            ll i;cin>>i;
            char ch;cin>>ch;
            st[s[i]-'a'].erase(i);
            st[ch-'a'].insert(i);
            s[i]=ch;
        }
        else{
            ll ans = 0 , l , r;
            cin>>l>>r;
            for(ll i=0;i<26;i++){
                auto pos = st[i].lower_bound(l);
                if(pos==st[i].end())continue;

                if( *pos <= r)ans++;
            }

            cout<<ans<<"\n";
        }
    }
}


