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;
typedef long long ll;
#define rep(k,i,n) for(ll i=k;i<n;++i)
int main(void){
    ll n,q,ans;
    string s;
    cin>>n>>s>>q;
    set<ll>se[26];
    rep(0,i,n)se[s[i]-'a'].insert(i+1);
    rep(0,nq,q){
        ll q0;
        cin>>q0;
        if(q0==1){
            ll i;
            char c;
            cin>>i>>c;
            rep(0,al,26)if(se[al].erase(i))break;
            se[c-'a'].insert(i);
        }else{
            ll l,r;
            cin>>l>>r;
            ans=0;
            rep(0,al,26){
                auto r0=se[al].lower_bound(l);
                if(r0!=se[al].end()&&*r0<r+1)++ans;
            }
            cout<<ans<<endl;
        }
    }
    
}
