Problem: 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.

Example

Input

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


Output

2
0
1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<long long> VL;
typedef vector<vector<long long>> VVL;
typedef pair<int,int> P;
typedef tuple<ll,ll,ll> tpl;

#define ALL(a)  (a).begin(),(a).end()
#define SORT(c) sort((c).begin(),(c).end())
#define REVERSE(c) reverse((c).begin(),(c).end())
#define EXIST(m,v) (m).find((v)) != (m).end()
#define LB(a,x) lower_bound((a).begin(), (a).end(), x) - (a).begin()
#define UB(a,x) upper_bound((a).begin(), (a).end(), x) - (a).begin()

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define RFOR(i,a,b) for(int i=(a)-1;i>=(b);--i)
#define RREP(i,n) RFOR(i,n,0)

#define en "\n"

constexpr double EPS = 1e-9;
constexpr double PI  = 3.1415926535897932;
constexpr int INF = 2147483647;
constexpr long long LINF = 1LL<<60;
constexpr long long MOD = 998244353;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

template <typename T,typename E>
struct LazySegmentTree{
    typedef function<T(T,T)> F;
    typedef function<T(T,E)> G;
    typedef function<E(E,E)> H;
    typedef function<E(E,ll)> P;
    int n;
    F f;
    G g;
    H h;
    P p;
    T d1;
    E d0;
    vector<T> val;
    vector<E> lazy;

    LazySegmentTree(vector<T> a, F f_, G g_, H h_, T d1_, E d0_, P p_=[](E a,int b){return a;}):
        f(f_),g(g_),h(h_),d1(d1_),d0(d0_),p(p_){

        int sz = a.size();
        n = 1;
        while (n < sz) n <<= 1;
        val.clear();
        val.resize(2 * n - 1, d1);
        lazy.clear();
        lazy.resize(2 * n - 1, d0);
        for (int i = 0; i < sz; i++) val[i + n - 1] = a[i];
        for (int i = n - 2; i >= 0; i--) val[i] = f(val[i * 2 + 1], val[i * 2 + 2]);
    }

    inline void eval(int k, int len){
        if(lazy[k] == d0) return;

        if(2*k+1 < 2*n-1){
            lazy[2*k+1] = h(lazy[2*k+1], lazy[k]);
            lazy[2*k+2] = h(lazy[2*k+2], lazy[k]);
        }
        val[k] = g(val[k], p(lazy[k], len));
        lazy[k] = d0;
    }

    void update(int a, int b, E x, int k=0, int l=0, int r=-1){
        // 0-indexed
        // [a,b)に対して操作
        if(r < 0) r = n;
        eval(k, r-l);
        if(r <=a || b <= l) return;

        if(a <= l && r <= b){
            lazy[k] = h(lazy[k], x);
            eval(k, r-l);
        }
        else{
            update(a, b, x, 2*k+1, l, (l+r)/2);
            update(a, b, x, 2*k+2, (l+r)/2, r);
            val[k] = f(val[2*k+1], val[2*k+2]);
        }
    }

    T query(int a, int b, int k=0, int l=0, int r=-1){
        // 0-indexed
        // [a,b)に対してquery
        if(r < 0) r = n;
        eval(k, r-l);
        if(r <= a || b <= l) return d1;
        if(a <= l && r <= b) return val[k];
        T val_l = query(a, b , 2*k+1, l, (l+r)/2);
        T val_r = query(a, b , 2*k+2, (l+r)/2, r);
        return f(val_l, val_r);
    }

    int findL(int a, int b, const function<bool(T)>& check, int k=0, int l=0, int r=-1){
        // 0-indexed
        // [a,b)でf(val[k])がtrueとなるような最小のkを返す 無ければ-1を返す
        // a      k         b
        // xxxxxxxoooooooooo

        if(r<0) r = n;
        eval(k, r-l);
        if(!check(val[k]) || (r<=a || b<=l)) return -1;
        if(r-l == 1) return l;

        int lv = findL(a, b, check, 2*k+1, l, (l+r)/2);
        if(lv != -1) return lv;

        int rv = findL(a, b, check, 2*k+2, (l+r)/2, r);
        return rv;
    }

    int findR(int a, int b, const function<bool(T)>& check, int k=0, int l=0, int r=-1){
        // 0-indexed
        // [a,b)でf(val[k])がtrueとなるような最大のkを返す 無ければ-1を返す
        // a      k         b
        // ooooooooxxxxxxxxx

        if(r<0) r = n;
        eval(k, r-l);
        if(!check(val[k]) || (r<=a || b<=l)) return -1;
        if(r-l == 1) return l;

        int rv = findR(a, b, check, 2*k+2, (l+r)/2, r);
        if(rv != -1) return rv;

        int lv = findR(a, b, check, 2*k+1, l, (l+r)/2);
        return lv;
    }
};

tpl f(tpl a, tpl b){
    ll at, a0, a1; tie(at, a0, a1) = a;
    ll bt, b0, b1; tie(bt, b0, b1) = b;
    return tpl(at+bt+a1*b0, a0+b0, a1+b1);
}
tpl g(tpl a, ll len){
    ll at, a0, a1; tie(at, a0, a1) = a;
    return tpl(a1*a0-at, a1, a0);
}
ll h(ll a, ll b){
    return a^b;
}
ll p(ll a, ll len){
    return a;
}

void Main(){
    int N,Q; cin >> N >> Q;
    vector<tpl> A(N);
    REP(i,N){
        int a; cin >> a;
        A[i] = a ? tpl(0,0,1) : tpl(0,1,0);
    }

    LazySegmentTree<tpl,ll> st(A,f,g,h,tpl(0,0,0),0,p);
    REP(_,Q){
        int T,l,r; cin >> T >> l >> r; l--;
        if(T == 1){
            st.update(l,r,1);
        }
        else{
            tpl t = st.query(l,r);
            ll ans = get<0>(t);
            cout << ans << en;
        }
    }
    return;
}

int main(void){
    cin.tie(0);cout.tie(0);ios_base::sync_with_stdio(0);cout<<fixed<<setprecision(15);
    int t=1; //cin>>t;
    REP(_,t) Main();
    return 0;
}