You are given a string s consisting of lowercase Latin letters and q queries for this string.

Recall that the substring s[l; r] of the string s is the string s_l s_{l + 1} ... s_r. For example, the substrings of "codeforces" are "code", "force", "f", "for", but not "coder" and "top".

There are two types of queries: 

  * 1~ pos~ c (1 ≤ pos ≤ |s|, c is lowercase Latin letter): replace s_{pos} with c (set s_{pos} := c); 
  * 2~ l~ r (1 ≤ l ≤ r ≤ |s|): calculate the number of distinct characters in the substring s[l; r]. 

Input

The first line of the input contains one string s consisting of no more than 10^5 lowercase Latin letters.

The second line of the input contains one integer q (1 ≤ q ≤ 10^5) — the number of queries.

The next q lines contain queries, one per line. Each query is given in the format described in the problem statement. It is guaranteed that there is at least one query of the second type.

Output

For each query of the second type print the answer for it — the number of distinct characters in the required substring in this query.

Examples

Input


abacaba
5
2 1 4
1 4 b
1 5 b
2 4 6
2 1 7


Output


3
1
2


Input


dfcbbcfeeedbaea
15
1 6 e
1 4 b
2 6 14
1 7 b
1 12 c
2 6 8
2 1 6
1 7 c
1 2 f
1 10 a
2 7 9
1 10 a
1 14 b
1 1 f
2 1 11


Output


5
2
5
2
6
import sys,math,itertools
from collections import Counter,deque,defaultdict
from bisect import bisect_left,bisect_right 
from heapq import heappop,heappush,heapify
from copy import deepcopy
mod = 10**9+7
INF = float('inf')
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
def inpl_1(): return list(map(lambda x:int(x)-1, sys.stdin.readline().split()))
def inps(): return sys.stdin.readline()
def inpsl(x): tmp = sys.stdin.readline(); return list(tmp[:x])
def err(x): print(x); exit()

import sys
input = sys.stdin.readline

def solve():
    ordA = ord('a')

    # N = int(input())
    Ss = input().rstrip()
    N = len(Ss)
    Q = int(input())
    querys = [tuple(input().split()) for _ in range(Q)]

    idEle = 0
    def _binOpe(x, y):
        return x | y
    def makeSegTree(numEle):
        numPow2 = 2 ** (numEle-1).bit_length()
        data = [idEle] * (2*numPow2)
        return data, numPow2
    def setInit(As):
        for iST, A in enumerate(As, numPow2):
            data[iST] = A
        for iST in reversed(range(1, numPow2)):
            data[iST] = _binOpe(data[2*iST], data[2*iST+1])
    def _update1(iA, A):
        iST = iA + numPow2
        data[iST] = A
    def update(iA, A):
        _update1(iA, A)
        iST = iA + numPow2
        while iST > 1:
            iST >>= 1
            data[iST] = _binOpe(data[2*iST], data[2*iST+1])
    def getValue(iSt, iEn):
        L = iSt + numPow2
        R = iEn + numPow2
        ans = idEle
        while L < R:
            if L & 1:
                ans = _binOpe(ans, data[L])
                L += 1
            if R & 1:
                R -= 1
                ans = _binOpe(ans, data[R])
            L >>= 1
            R >>= 1
        return ans

    data, numPow2 = makeSegTree(N)
    As = [(1<<(ord(S)-ordA)) for S in Ss]
    setInit(As)

    anss = []
    for tp, *vs in querys:
        if tp == '1':
            i, c = vs
            i = int(i)
            update(i-1, 1<<(ord(c)-ordA))
        else:
            L, R = vs
            L, R = int(L), int(R)
            v = getValue(L-1, R)
            anss.append(bin(v).count('1'))

    print('\n'.join(map(str, anss)))


solve()
