Problem:
Score : 100 points

Problem Statement

You are given a binary array A=(A_1,A_2,\cdots,A_N) of length N.

Process Q queries of the following types. The i-th query is represented by three integers T_i,L_i,R_i.

T_i=1: Replace the value of A_j with 1-A_j for each L_i \leq j \leq R_i.

T_i=2: Calculate the inversion(*) of the array A_{L_i},A_{L_i+1},\cdots,A_{R_i}．

Note：The inversion of the array x_1,x_2,\cdots,x_k is the number of the pair of integers i,j with 1 \leq i < j \leq k, x_i > x_j.

Constraints

1 \leq N \leq 2 \times 10^5

0 \leq A_i \leq 1

1 \leq Q \leq 2 \times 10^5

1 \leq T_i \leq 2

1 \leq L_i \leq R_i \leq N

All values in Input are integer.

Input

Input is given from Standard Input in the following format:

N Q
A_1 A_2 \cdots A_N
T_1 L_1 R_1
T_2 L_2 R_2
\vdots
T_Q L_Q R_Q

Output

For each query with T_i=2, print the answer.

Sample Input 1

5 5
0 1 0 0 1
2 1 5
1 3 4
2 2 5
1 1 3
2 1 2

Sample Output 1

2
0
1

First query: Print 2, which is the inversion of (A_1,A_2,A_3,A_4,A_5)=(0,1,0,0,1).

Second query: Replace the value of A_3 and A_4 with 1 and 1, respectively.

Third query: Print 0, which is the inversion of (A_2,A_3,A_4,A_5)=(1,1,1,1).

Fourth query: Replace the value of A_1, A_2 and A_4 with 1, 0 and 0, respectively.

Fifth query: Print 1, which is the inversion of (A_1,A_2)=(1,0).

Solution in Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;

class LazySegTree<S, F> {
    final int MAX;

    final int N;
    final int Log;
    final java.util.function.BinaryOperator<S> Op;
    final S E;
    final java.util.function.BiFunction<F, S, S> Mapping;
    final java.util.function.BinaryOperator<F> Composition;
    final F Id;

    final S[] Dat;
    final F[] Laz;

    @SuppressWarnings("unchecked")
    public LazySegTree(int n, java.util.function.BinaryOperator<S> op, S e, java.util.function.BiFunction<F, S, S> mapping, java.util.function.BinaryOperator<F> composition, F id) {
        this.MAX = n;
        int k = 1;
        while (k < n) k <<= 1;
        this.N = k;
        this.Log = Integer.numberOfTrailingZeros(N);
        this.Op = op;
        this.E = e;
        this.Mapping = mapping;
        this.Composition = composition;
        this.Id = id;
        this.Dat = (S[]) new Object[N << 1];
        this.Laz = (F[]) new Object[N];
        java.util.Arrays.fill(Dat, E);
        java.util.Arrays.fill(Laz, Id);
    }

    public LazySegTree(S[] dat, java.util.function.BinaryOperator<S> op, S e, java.util.function.BiFunction<F, S, S> mapping, java.util.function.BinaryOperator<F> composition, F id) {
        this(dat.length, op, e, mapping, composition, id);
        build(dat);
    }

    private void build(S[] dat) {
        int l = dat.length;
        System.arraycopy(dat, 0, Dat, N, l);
        for (int i = N - 1; i > 0; i--) {
            Dat[i] = Op.apply(Dat[i << 1 | 0], Dat[i << 1 | 1]);
        }
    }

    private void push(int k) {
        if (Laz[k] == Id) return;
        int lk = k << 1 | 0, rk = k << 1 | 1;
        Dat[lk] = Mapping.apply(Laz[k], Dat[lk]);
        Dat[rk] = Mapping.apply(Laz[k], Dat[rk]);
        if (lk < N) Laz[lk] = Composition.apply(Laz[k], Laz[lk]);
        if (rk < N) Laz[rk] = Composition.apply(Laz[k], Laz[rk]);
        Laz[k] = Id;
    }

    private void pushTo(int k) {
        for (int i = Log; i > 0; i--) push(k >> i);
    }

    private void pushTo(int lk, int rk) {
        for (int i = Log; i > 0; i--) {
            if (((lk >> i) << i) != lk) push(lk >> i);
            if (((rk >> i) << i) != rk) push(rk >> i);
        }
    }

    private void updateFrom(int k) {
        k >>= 1;
        while (k > 0) {
            Dat[k] = Op.apply(Dat[k << 1 | 0], Dat[k << 1 | 1]);
            k >>= 1;
        }
    }

    private void updateFrom(int lk, int rk) {
        for (int i = 1; i <= Log; i++) {
            if (((lk >> i) << i) != lk) {
                int lki = lk >> i;
                Dat[lki] = Op.apply(Dat[lki << 1 | 0], Dat[lki << 1 | 1]);
            }
            if (((rk >> i) << i) != rk) {
                int rki = (rk - 1) >> i;
                Dat[rki] = Op.apply(Dat[rki << 1 | 0], Dat[rki << 1 | 1]);
            }
        }
    }

    public void set(int p, S x) {
        exclusiveRangeCheck(p);
        p += N;
        pushTo(p);
        Dat[p] = x;
        updateFrom(p);
    }

    public S get(int p) {
        exclusiveRangeCheck(p);
        p += N;
        pushTo(p);
        return Dat[p];
    }

    public S prod(int l, int r) {
        if (l > r) {
            throw new IllegalArgumentException(
                String.format("Invalid range: [%d, %d)", l, r)
            );
        }
        inclusiveRangeCheck(l);
        inclusiveRangeCheck(r);
        if (l == r) return E;
        l += N; r += N;
        pushTo(l, r);
        S sumLeft = E, sumRight = E;
        while (l < r) {
            if ((l & 1) == 1) sumLeft = Op.apply(sumLeft, Dat[l++]);
            if ((r & 1) == 1) sumRight = Op.apply(Dat[--r], sumRight);
            l >>= 1; r >>= 1;
        }
        return Op.apply(sumLeft, sumRight);
    }

    public S allProd() {
        return Dat[1];
    }

    public void apply(int p, F f) {
        exclusiveRangeCheck(p);
        p += N;
        pushTo(p);
        Dat[p] = Mapping.apply(f, Dat[p]);
        updateFrom(p);
    }

    public void apply(int l, int r, F f) {
        if (l > r) {
            throw new IllegalArgumentException(
                String.format("Invalid range: [%d, %d)", l, r)
            );
        }
        inclusiveRangeCheck(l);
        inclusiveRangeCheck(r);
        if (l == r) return;
        l += N; r += N;
        pushTo(l, r);
        for (int l2 = l, r2 = r; l2 < r2;) {
            if ((l2 & 1) == 1) {
                Dat[l2] = Mapping.apply(f, Dat[l2]);
                if (l2 < N) Laz[l2] = Composition.apply(f, Laz[l2]);
                l2++;
            }
            if ((r2 & 1) == 1) {
                r2--;
                Dat[r2] = Mapping.apply(f, Dat[r2]);
                if (r2 < N) Laz[r2] = Composition.apply(f, Laz[r2]);
            }
            l2 >>= 1; r2 >>= 1;
        }
        updateFrom(l, r);
    }

    public int maxRight(int l, java.util.function.Predicate<S> g) {
        inclusiveRangeCheck(l);
        if (!g.test(E)) {
            throw new IllegalArgumentException("Identity element must satisfy the condition.");
        }
        if (l == MAX) return MAX;
        l += N;
        pushTo(l);
        S sum = E;
        do {
            l >>= Integer.numberOfTrailingZeros(l);
            if (!g.test(Op.apply(sum, Dat[l]))) {
                while (l < N) {
                    push(l);
                    l = l << 1;
                    if (g.test(Op.apply(sum, Dat[l]))) {
                        sum = Op.apply(sum, Dat[l]);
                        l++;
                    }
                }
                return l - N;
            }
            sum = Op.apply(sum, Dat[l]);
            l++;
        } while ((l & -l) != l);
        return MAX;
    }

    public int minLeft(int r, java.util.function.Predicate<S> g) {
        inclusiveRangeCheck(r);
        if (!g.test(E)) {
            throw new IllegalArgumentException("Identity element must satisfy the condition.");
        }
        if (r == 0) return 0;
        r += N;
        pushTo(r - 1);
        S sum = E;
        do {
            r--;
            while (r > 1 && (r & 1) == 1) r >>= 1;
            if (!g.test(Op.apply(Dat[r], sum))) {
                while (r < N) {
                    push(r);
                    r = r << 1 | 1;
                    if (g.test(Op.apply(Dat[r], sum))) {
                        sum = Op.apply(Dat[r], sum);
                        r--;
                    }
                }
                return r + 1 - N;
            }
            sum = Op.apply(Dat[r], sum);
        } while ((r & -r) != r);
        return 0;
    }

    private void exclusiveRangeCheck(int p) {
        if (p < 0 || p >= MAX) {
            throw new IndexOutOfBoundsException(
                String.format("Index %d is not in [%d, %d).", p, 0, MAX)
            );
        }
    }

    private void inclusiveRangeCheck(int p) {
        if (p < 0 || p > MAX) {
            throw new IndexOutOfBoundsException(
                String.format("Index %d is not in [%d, %d].", p, 0, MAX)
            );
        }
    }
}
public class Main {
    private static class S{
        long zero;
        long one;
        long inv;
        public S(long zero, long one, long inv){
            this.zero = zero;
            this.one = one;
            this.inv = inv;
        }
        S swap01(){
            return new S(this.one, this.zero, this.zero*this.one-this.inv);
        }

        static S connect(S a, S b){
            return new S(a.zero+b.zero, a.one+b.one, a.inv+b.inv+a.one*b.zero);
        }
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int Q = sc.nextInt();

        S[] A = new S[N+1];
        A[0] = new S(0,0,0);
        for(int n=1; n<=N; n++) A[n] = sc.nextLong()==0L ? new S(1,0,0) : new S(0,1,0);

        LazySegTree<S, Boolean> seg = new LazySegTree<S,Boolean>(A, S::connect, new S(0,0,0), (l,r) -> l ? r.swap01() : r, (x,y)->x^y, false);

        for(int q=0; q<Q; q++){
            int t = sc.nextInt();
            if(t==1){
                int l = sc.nextInt();
                int r = sc.nextInt();
                seg.apply(l, r+1, true);
            }else if(t==2){
                int l = sc.nextInt();
                int r = sc.nextInt();
                System.out.println(seg.prod(l, r+1).inv);
            }
        }
        
    }
}