count minimum number of possible anagrams after atmost k change (IMPORTANT OA)

https://practice.geeksforgeeks.org/contest/gfg-weekly-coding-contest-86/problems/#

Given a string s of length n, maximize the number of possible *anagrams by doing atmost k special operations.
A special operation is defined as replacing any character with any of the available 26 lower case alphabets. After doing atmost k operations return the total number of anagrams.
Since the answer can be very large, return the answer modulo 10^9+7.
*anagrams :- An anagram is a word or phrase formed by rearranging the letters of another word or phrase. For example, the word "listen" can be rearranged into "silent".

Input : N=3  K=1  s="aac"
Output : 6
Explaination :
After applying one operation we can "a" on first index to "b" and
6 possible anagrams are "abc","acb","bca","bac","cab","cba"

Input : N=3   K=2   s="abc"
Output: 6
Explaination :
The string "abc" itself has the maximum possible anagrams.

Your Task:
Your task is to count the number of possible anagrams modulo 10^9 + 7 .

Constraints:
1 <= N <= 10 ^ 5
1 <= K <= 10 ^ 5

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    int mod=1000000007;
    long long int binaryExpo(long long int x, long long int power){
        long long int res=1;
        while(power>0){
            if(power%2==0){
                x=(x*x)%mod;
                power/=2;
            }
            else{
                res=(res*x)%mod;
                power=power-1;
            }
        }
        return res;
    }

    int fact(long long int n){
        long long int res=1;
        for(int i=1;i<=n;i++){
            res=(res*i)%mod;
        }
        return res%mod;
    }

    int maximumPossible(int n,int k,string s){
        int freq[26]={0};
        for(auto c: s){
            freq[c-'a']++;
        }

        multiset<pair<int,char>> mpp; // you can get away using only freq, but using a char helps to visual.

        for(int i=0;i<26;i++){
            mpp.insert({freq[i],i+'a'});
        }

        k=min(k,n);  // the maximum change you can do is just n only at max

        while(k>0 && abs(mpp.begin()->first-mpp.rbegin()->first)>=2){
            auto tp=*mpp.begin();
            mpp.erase(mpp.begin());

            auto rp=*mpp.rbegin();
            mpp.erase(--mpp.end());

            tp.first++;
            mpp.insert(tp);

            rp.first--;
            mpp.insert(rp);

            k--;
        }

        int res=fact(n);

        for(auto m: mpp){
          int fx=fact(m.first);
          res=(res*binaryExpo(fx,mod-2))%mod;  // fermat little theorem
        }
        return res;
    }