#include <iostream>
#include <stdio.h>
#define lc (o << 1)
#define rc (o << 1 | 1)
using namespace std;
typedef long long LL;
#define S (1e5 + 5)
double sum1[400005], sum2[400005], tag[400005];
double a[100005];
int n, q;
void pushup(int o) {
	sum1[o] = sum1[lc] + sum1[rc];
	sum2[o] = sum2[lc] + sum2[rc];
}
void pushdown(int o, int l, int r) {
	tag[lc] += tag[o];
	tag[rc] += tag[o];
	int mid = (l + r) / 2;
	sum2[lc] += 2 * tag[o] * sum1[lc] + (mid - l + 1) * tag[o] * tag[o];
	sum2[rc] += 2 * tag[o] * sum1[rc] + (r - mid) * tag[o] * tag[o];
	sum1[lc] += (mid - l + 1) * tag[o];
	sum1[rc] += (r - mid) * tag[o];
	tag[o] = 0;
}
void build(int o, int l, int r) {
	if (l == r) {
		sum1[o] = a[l];
		sum2[o] = a[l] * a[l];
		return;
	}
	int mid = (l + r) / 2;
	build(lc, l, mid);
	build(rc, mid + 1, r);
	pushup(o);
}
double query1(int o, int l, int r, int ql, int qr) {
	if (ql <= l && r <= qr) {
		return sum1[o];
	}
	int mid = (l + r) / 2;
	double ans = 0;
	pushdown(o, l, r);
	if (ql <= mid) {
		ans += query1(lc, l, mid, ql, qr);
	}
	if (mid < qr) {
		ans += query1(rc, mid + 1, r, ql, qr);
	}
	return ans;
}
double query2(int o, int l, int r, int ql, int qr) {
	if (ql <= l && r <= qr) {
		return sum2[o];
	}
	int mid = (l + r) / 2;
	double ans = 0;
	pushdown(o, l, r);
	if (ql <= mid) {
		ans += query2(lc, l, mid, ql, qr);
	}
	if (mid < qr) {
		ans += query2(rc, mid + 1, r, ql, qr);
	}
	return ans;
}
void update(int o, int l, int r, int pl, int pr, int v) {
	if (pl <= l && r <= pr) {
		tag[o] += v;
		sum2[o] += 2 * v * sum1[o] + (r - l + 1) * v * v;
		sum1[o] += (r - l + 1) * v;
		return;
	}
	if (r < pl || l > pr) {
		return;
	}
	pushdown(o, l, r);
	int mid = (l + r) / 2;
	update(lc, l, mid, pl, pr, v);
	update(rc, mid + 1, r, pl, pr, v);
	pushup(o);
}
int main() {
	cin >> n >> q;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	build(1, 1, n);
	int op, x, y;
	double k;
	while (q--) {
		cin >> op >> x >> y;
		if (op == 2) {
			printf("%.4f\n", query1(1, 1, n, x, y) / (y - x + 1));
		}
		if (op == 1) {
			cin >> k;
			update(1, 1, n, x, y, k);
		} else {
			double s1 = query2(1, 1, n, x, y) / (y - x + 1);
			double s2 = query1(1, 1, n, x, y) / (y - x + 1);
			printf("%.4f\n", s1 - s2 * s2);
		}
	}
	return 0;
}

Orz