#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>
using namespace std;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
using ll = long long;
const int inf=1e9+7;
const ll longinf=1LL<<60 ;
const ll mod=1e9+7 ;


template<typename T>
struct SegmentTree{
private:
    int n;
    T E;
    vector<T> node;
    inline void updatef(T& x,T& y){
        x = y;
    }
    inline T queryf(T& x,T& y){
        return {x.first + y.first, x.second+y.second};
    }

public:
    SegmentTree(int sz,T E_):E(E_){
        n=1;
        while(n<sz)n<<=1;
        node.resize(2*n-1,E);
    }

    SegmentTree(vector<T>& A,T E_):E(E_){
        int sz=A.size();
        n=1;
        while(n<sz)n<<=1;
        node.resize(2*n-1,E);
        rep(i,sz)node[i+n-1]=A[i];
        for(int i=n-2;i>=0;--i){
            node[i]=queryf(node[2*i+1], node[2*i+2]);
        }
    }
    void update(int k,T x){
        k+=n-1;
        updatef(node[k],x);
        while(k>0){
            k=(k-1)/2;
            node[k]=queryf(node[2*k+1], node[2*k+2]);
        }
    }
       //[a,b)での和を返す
    T get(int a,int b,int k=0,int l=0,int r=-1){
        if(r<0)r=n;
        if(r<=a||b<=l)return E;
        if(a<=l&&r<=b)return node[k];
        T xl=get(a,b,2*k+1,l,(l+r)/2);
        T xr=get(a,b,2*k+2,(l+r)/2,r);
        return queryf(xl, xr);
    }
};
int main(){
    int n,q;
    cin>>n>>q;
    SegmentTree<pair<ll,ll>> sg(n,{0,0});
    vector<ll> x(n+q);
    rep(i,n)cin>>x[i];
    vector<int> l(q), r(q);
    rep(i,q){
        cin>>l[i]>>l[i]>>r[i]>>x[n+i];
        --l[i];
    }
    vector<int> ord(n+q);
    vector<ll> ans(q);
    iota(ord.begin(),ord.end(),0);
    sort(ord.begin(),ord.end(),[&](int i, int j){
        return x[i]>x[j];
    });
    for(auto i : ord){
        if(i<n){
            sg.update(i,{x[i],1});
        }
        else {
            auto ret = sg.get(l[i-n],r[i-n]);
            ans[i-n]=ret.first-ret.second*x[i];
        }
    }
    rep(i,q)cout<<ans[i]<<endl;
    return 0;
}