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 C++ (GCC 9.2.1 with AC Library v1.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

#include <atcoder/all>

using namespace std;
using namespace atcoder;

using ll = long long;
using pii = pair<int, int>;
template <class T>
using V = vector<T>;
template <class T>
using VV = V<V<T>>;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i, n) rep2(i, 0, n)
#define rep2(i, m, n) for (int i = m; i < (n); i++)
#define per(i, b) per2(i, 0, b)
#define per2(i, a, b) for (int i = int(b) - 1; i >= int(a); i--)
#define ALL(c) (c).begin(), (c).end()
#define SZ(x) ((int)(x).size())

constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n - 1); }

template <class T, class U>
void chmin(T& t, const U& u) {
    if (t > u) t = u;
}
template <class T, class U>
void chmax(T& t, const U& u) {
    if (t < u) t = u;
}

template <class T, class U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
    os << "(" << p.first << "," << p.second << ")";
    return os;
}

template <class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
    os << "{";
    rep(i, v.size()) {
        if (i) os << ",";
        os << v[i];
    }
    os << "}";
    return os;
}

#ifdef LOCAL
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
    cerr << " " << H;
    debug_out(T...);
}
#define debug(...) \
    cerr << __LINE__ << " [" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#define dump(x) cerr << __LINE__ << " " << #x << " = " << (x) << endl
#else
#define debug(...) (void(0))
#define dump(x) (void(0))
#endif

struct S {
    ll c0, c1, cinv;
};

S op(S l, S r) {
    return S{l.c0 + r.c0, l.c1 + r.c1, l.cinv + r.cinv + l.c1 * r.c0};
}

S e() { return S{0ll, 0ll, 0ll}; }

using F = bool;

S mapping(F l, S r) {
    if (l) {
        return S{r.c1, r.c0, r.c0 * r.c1 - r.cinv};
    } else {
        return r;
    }
}

F composition(F l, F r) { return l ^ r; }
F id() { return false; }

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int N, Q;
    cin >> N >> Q;
    V<int> A(N);
    V<S> init_A(N);
    rep(i, N) {
        cin >> A[i];
        init_A[i] = S{!A[i], A[i], 0};
    }

    lazy_segtree<S, op, e, F, mapping, composition, id> seg(init_A);
    while (Q--) {
        int t, l, r;
        cin >> t >> l >> r;
        --l;
        if (t == 1) {
            seg.apply(l, r, true);
        } else {
            auto res = seg.prod(l, r);
            cout << res.cinv << '\n';
        }
    }

    return 0;
}