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;
#define rep(i,n) for(ll i=0;i<n;i++)
#define repl(i,l,r) for(ll i=(l);i<(r);i++)
#define per(i,n) for(ll i=n-1;i>=0;i--)
#define perl(i,r,l) for(ll i=r-1;i>=l;i--)
#define fi first
#define se second
#define pb push_back
#define ins insert
#define pqueue(x) priority_queue<x,vector<x>,greater<x>>
#define all(x) (x).begin(),(x).end()
#define CST(x) cout<<fixed<<setprecision(x)
#define vtpl(x,y,z) vector<tuple<x,y,z>>
#define rev(x) reverse(x);
using ll=long long;
using vl=vector<ll>;
using vvl=vector<vector<ll>>;
using pl=pair<ll,ll>;
using vpl=vector<pl>;
using vvpl=vector<vpl>;
const ll MOD=1000000007;
const ll MOD9=998244353;
const int inf=1e9+10;
const ll INF=4e18;
const ll dy[8]={1,0,-1,0,1,1,-1,-1};
const ll dx[8]={0,-1,0,1,1,-1,1,-1};
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a <= b) {
        a = b;
        return true;
    }
    return false;
}


template <typename X, typename M>
struct SegTreeLazy {//遅延セグ木　単位元に注意(updateなら選ばれない数、affineなら(1,0))
    using FX = function<X(X, X)>;
    using FA = function<X(X, M, int)>;
    using FM = function<M(M, M)>;
    int n;
    FX fx;
    FA fa;
    FM fm;
    const X ex;
    const M em;
    vector<X> dat;
    vector<M> lazy;
    SegTreeLazy(int n_, FX fx_, FA fa_, FM fm_, X ex_, M em_)
        : n(), fx(fx_), fa(fa_), fm(fm_), ex(ex_), em(em_), dat(n_ * 4, ex), lazy(n_ * 4, em) {
        int x = 1;
        while (n_ > x) x *= 2;
        n = x;
    }
    void set(int i, X x) { dat[i + n - 1] = x; }
    void build() {
        for (int k = n - 2; k >= 0; k--) dat[k] = fx(dat[2 * k + 1], dat[2 * k + 2]);
    }
    /* lazy eval */
    void eval(int k, int len) {
        if (lazy[k] == em) return;  // 更新するものが無ければ終了
        if (k < n - 1) {            // 葉でなければ子に伝搬
            lazy[k * 2 + 1] = fm(lazy[k * 2 + 1], lazy[k]);
            lazy[k * 2 + 2] = fm(lazy[k * 2 + 2], lazy[k]);
        }
        // 自身を更新
        dat[k] = fa(dat[k],lazy[k],len);//fa(dat[k], fp(lazy[k], len));
        lazy[k] = em;
    }
    void update(int a, int b, M x, int k, int l, int r) {
        eval(k, r - l);
        if (a <= l && r <= b) {  // 完全に内側の時
            lazy[k] = fm(lazy[k], x);
            eval(k, r - l);
        } else if (a < r && l < b) {                     // 一部区間が被る時
            update(a, b, x, k * 2 + 1, l, (l + r) / 2);  // 左の子
            update(a, b, x, k * 2 + 2, (l + r) / 2, r);  // 右の子
            dat[k] = fx(dat[k * 2 + 1], dat[k * 2 + 2]);
        }
    }
    void update(int a, int b, M x) { update(a, b, x, 0, 0, n); }
    X query_sub(int a, int b, int k, int l, int r) {
        eval(k, r - l);
        if (r <= a || b <= l) {  // 完全に外側の時
            return ex;
        } else if (a <= l && r <= b) {  // 完全に内側の時
            return dat[k];
        } else {  // 一部区間が被る時
            X vl = query_sub(a, b, k * 2 + 1, l, (l + r) / 2);
            X vr = query_sub(a, b, k * 2 + 2, (l + r) / 2, r);
            return fx(vl, vr);
        }
    }
    X query(int a, int b) { return query_sub(a, b, 0, 0, n); }
    X operator[](int i){
        return query(i,i+1);
    }
};

const int mod = MOD9;
const int max_n = 200005;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }
  bool operator==(const mint &p) const { return x == p.x; }
  bool operator!=(const mint &p) const { return x != p.x; }
 
  // for prime mod
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}

struct node{
    ll a,b,rev;
};
int main(){
    ll n,q;cin >> n >>q;
    auto fx=[](node a,node b){
        return (node){a.a+b.a,a.b+b.b,a.rev+b.rev+(a.b*b.a)};
    };
    auto fa=[](node a,bool b,int len){
        if(b)return (node){a.b,a.a,a.a*a.b-a.rev};
        else return a;
    };
    auto fm=[](bool a,bool b){return a^b;};
    SegTreeLazy<node,bool> st(n,fx,fa,fm,(node){0,0,0},false);
    rep(i,n){
        ll a;cin >> a;
        if(a)st.set(i,{0,1,0});
        else st.set(i,{1,0,0});
    }
    st.build();
    while(q--){
        ll t,l,r;cin >> t >> l >>r;
        if(t==1){
            st.update(l-1,r,true);
        }
        else{
            cout << st.query(l-1,r).rev <<endl;
        }
    }
}