//基本的にはa[l]+…+a[r] - x*(r-l+1)が答えだが、Σ(x-a[i]) (a[i] < x)だけ余分に答えが引かれてしまう。
//よって、a[i] < xなるi(l≦i≦r)の個数、a[i] < xなるi(l≦i≦r)についてa[i]の和、を知りたい。
//これは領域木でもできるけど実装が面倒くさいのと計算量がlog^2Nなのは嫌いなので、ソートしてセグ木で殴ります。
#include <iostream>
#include <vector>
#include <tuple>
#include <queue>
#include <algorithm>
#include <functional>
#define int long long
#define rep(i, n) for(i = 0; i < n; i++)
using namespace std;

const int DEPTH = 17;
struct SegTree {
	int d[1 << (DEPTH + 1)];
	
	SegTree() {
		int i;
		rep(i, 1 << (DEPTH + 1)) { d[i] = 0; }
	}
	void add(int pos, int x) {
		pos += (1 << DEPTH) - 1;
		d[pos] += x;
		while (pos > 0) {
			pos = (pos - 1) / 2;
			d[pos] = d[pos * 2 + 1] + d[pos * 2 + 2];
		}
	}
	int sum(int l, int r, int a = 0, int b = (1 << DEPTH), int id = 0) {
		if (a >= r || b <= l) return 0;
		if (l <= a && b <= r) return d[id];
		int x = sum(l, r, a, (a + b) / 2, id * 2 + 1);
		int y = sum(l, r, (a + b) / 2, b, id * 2 + 2);
		return x + y;
	}
};

typedef pair<int, int> P;
typedef tuple<int, int, int, int> T;	//(x, l, r, id)
SegTree seg;		//まあ累積和でいいんですが
SegTree segCnt;
SegTree segSum;
int n, q;
int a[100000];
priority_queue<P, vector<P>, greater<P>> que;
T query[100000];
int ans[100000];

signed main() {
	int i;
	
	cin >> n >> q;
	rep(i, n) cin >> a[i];
	rep(i, n) que.push(P(a[i], i));
	rep(i, n) seg.add(i, a[i]);
	rep(i, q) {
		int t, l, r, x;
		cin >> t >> l >> r >> x;
		l--; r--;
		query[i] = T(x, l, r, i);
	}
	sort(query, query + q);
	
	rep(i, q) {
		int x = get<0>(query[i]);
		int l = get<1>(query[i]);
		int r = get<2>(query[i]);
		int id = get<3>(query[i]);
		
		while (!que.empty()) {
			P now = que.top();
			if (now.first >= x) break;
			int index = now.second;
			segCnt.add(index, 1);
			segSum.add(index, a[index]);
			que.pop();
		}
		
		int allSum = seg.sum(l, r + 1);
		int smallCnt = segCnt.sum(l, r + 1);
		int smallSum = segSum.sum(l, r + 1);
		
		ans[id] = allSum - x * (r + 1 - l - smallCnt) - smallSum;
	}
	
	rep(i, q) cout << ans[i] << endl;
	return 0;
}