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;
int main(){
int N, i, j; string S; cin >> N >> S; 
vector< set<int> > al(26);
for(i=0; i<26; i++){ al[i]={N}; }
for(i=0; i<N; i++){ al[S[i]-'a'].insert(i); } 
int Q; cin >> Q;
for(i=0; i<Q; i++){
  int a; cin >> a;
  if(a==1){ 
    int k; char c; cin >> k >> c; k--;
    if(S[k]!=c){ al[S[k]-'a'].erase(k); S[k]=c; al[c-'a'].insert(k); }
  }else{
    int l, r, ans=0; cin >> l >> r; l--; r--;
    for(j=0; j<26; j++){ if(*al[j].lower_bound(l)<=r){ ans++; } }
    cout << ans << "\n";
  }
}
}