#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=2e5+10,M=1010;
int a[N],s[M],mx[M],n,B;
void upd(int l,int r)
{
	int bl=(l-1)/B+1,br=(r-1)/B+1;
	if(bl==br)
	{
		for(int i=l;i<=r;i++)
			s[bl]-=a[i],a[i]=sqrt(a[i]),s[bl]+=a[i];
		mx[bl]=0;
		for(int i=(bl-1)*B+1;i<=min(n,bl*B);i++)
			mx[bl]=max(mx[bl],a[i]);
	}
	else
	{
		for(int i=l;i<=bl*B;i++)
			s[bl]-=a[i],a[i]=sqrt(a[i]),s[bl]+=a[i];
		mx[bl]=0;
		for(int i=(bl-1)*B+1;i<=bl*B;i++)
			mx[bl]=max(mx[bl],a[i]);
		for(int i=(br-1)*B+1;i<=r;i++)
			s[br]-=a[i],a[i]=sqrt(a[i]),s[br]+=a[i];
		mx[br]=0;
		for(int i=(br-1)*B+1;i<=min(n,br*B);i++)
			mx[br]=max(mx[br],a[i]);
		for(int i=bl+1;i<br;i++)if(mx[i]>1)
		{
			for(int j=(i-1)*B+1;j<=i*B;j++)
				s[i]-=a[j],a[j]=sqrt(a[j]),s[i]+=a[j];
			mx[i]=0;
			for(int j=(i-1)*B+1;j<=i*B;j++)
				mx[i]=max(mx[i],a[j]);
		}
	}
}
int query(int l,int r)
{
	int bl=(l-1)/B+1,br=(r-1)/B+1,ans=0;
	if(bl==br)
	{
		for(int i=l;i<=r;i++)
			ans+=a[i];
	}
	else
	{
		for(int i=l;i<=bl*B;i++)ans+=a[i];
		for(int i=(br-1)*B+1;i<=r;i++)ans+=a[i];
		for(int i=bl+1;i<br;i++)ans+=s[i];
	}
	return ans;
}
signed main()
{
	cin>>n;B=sqrt(n);
	for(int i=1;i<=n;i++)cin>>a[i],s[(i-1)/B+1]+=a[i],mx[(i-1)/B+1]=max(mx[(i-1)/B+1],a[i]);
	for(int i=1;i<=n;i++)
	{
		int op,l,r,c;cin>>op>>l>>r>>c;
		if(op==0)upd(l,r);
		else cout<<query(l,r)<<'\n';
	}
	return 0;
}

#include <bits/stdc++.h>
#define int long long
using namespace std;

const int N = 5e4 + 10, sqrtN = 250;
// 区间开方，区间求和
// a[i]记录每个点的值，b[i]记录每个点i所在块;
// sum[i]记录第i个块的元素和，mx[i]记录第i个块的最大值
// 每个块i的左端点L[i]、右端点R[i]
int n, a[N], b[N], L[sqrtN], R[sqrtN], sum[sqrtN], mx[sqrtN];

signed main() {
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    int B = sqrt(n), cnt = (n + B - 1) / B; // B为每块的长度，cnt为总块数
    for (int i = 1; i <= n; i++) {
        b[i] = (i - 1) / B + 1;
    }
    memset(sum, 0, sizeof(sum));
    memset(mx, 0, sizeof(mx));
    for (int i = 1; i <= cnt; i++) {
        L[i] = (i - 1) * B + 1;
        R[i] = min(i * B, n);
        for (int j = L[i]; j <= R[i]; j++) {
            sum[i] += a[j];
            mx[i] = max(mx[i], a[j]);
        }
    }
    for (int i = 1; i <= n; i++) {
        int op, l, r, c;
        cin >> op >> l >> r >> c;
        if (op == 0) {
            if (b[l] == b[r]) { // 如果l和r在同一块内
                if (mx[b[l]] <= 1) continue; // 块内最大值<=1，无需操作
                for (int j = l; j <= r; j++) {
                    sum[b[l]] -= a[j];
                    a[j] = sqrt(a[j]);
                    sum[b[l]] += a[j];
                }
                mx[b[l]] = 0;
                for (int j = L[b[l]]; j <= R[b[l]]; j++)
                    mx[b[l]] = max(mx[b[l]], a[j]);
            } else {
                if (mx[b[l]] > 1) {
                    for (int j = l; j <= R[b[l]]; j++) {
                        sum[b[l]] -= a[j];
                        a[j] = sqrt(a[j]);
                        sum[b[l]] += a[j];
                    }
                    mx[b[l]] = 0;
                    for (int j = L[b[l]]; j <= R[b[l]]; j++)
                        mx[b[l]] = max(mx[b[l]], a[j]);
                }
                for (int j = b[l] + 1; j <= b[r] - 1; j++) {
                    if (mx[j] <= 1) continue; // 块内最大值<=1，无需操作
                    for (int k = L[j]; k <= R[j]; k++) {
                        sum[j] -= a[k];
                        a[k] = sqrt(a[k]);
                        sum[j] += a[k];
                    }
                    mx[j] = 0;
                    for (int k = L[j]; k <= R[j]; k++)
                        mx[j] = max(mx[j], a[k]);
                }
                if (mx[b[r]] > 1) {
                    for (int j = L[b[r]]; j <= r; j++) {
                        sum[b[r]] -= a[j];
                        a[j] = sqrt(a[j]);
                        sum[b[r]] += a[j];
                    }
                    mx[b[r]] = 0;
                    for (int j = L[b[r]]; j <= R[b[r]]; j++)
                        mx[b[r]] = max(mx[b[r]], a[j]);
                }
            }
        } else {
            int ans = 0;
            if (b[l] == b[r]) { // 如果l和r在同一块内
                for (int j = l; j <= r; j++)
                    ans += a[j];
            } else {
                for (int j = l; j <= R[b[l]]; j++)
                    ans += a[j];
                for (int j = b[l] + 1; j <= b[r] - 1; j++)
                    ans += sum[j];
                for (int j = L[b[r]]; j <= r; j++)
                    ans += a[j];
            }
            cout << ans << '\n';
        }
    }
    return 0;
}