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

const int N = 5e4 + 10, sqrtN = 250;

int n, a[N], b[N], c[N], L[sqrtN], R[sqrtN], tag[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];
        c[i] = a[i];
    }
    int B = sqrt(n), cnt = (n + B - 1) / B;
    for (int i = 1; i <= n; i++) b[i] = (i - 1) / B + 1;
    for (int i = 1; i <= cnt; i++) {
        L[i] = (i - 1) * B + 1;
        R[i] = min(i * B, n);
        sort(c + L[i], c + R[i] + 1);
    }
	memset(tag, 0, sizeof(tag));
    for (int i = 1; i <= n; i++) {
        int op, l, r, v;
        cin >> op >> l >> r >> v;
        if (op == 0) {
            if (b[l] == b[r]) {

                for (int i = l; i <= r; i++) a[i] += v;
                for (int i = L[b[l]]; i <= R[b[l]]; i++) c[i] = a[i];
                sort(c + L[b[l]], c + R[b[l]] + 1);

            } else {

                for (int i = l; i <= R[b[l]]; i++) a[i] += v;
                for (int i = L[b[l]]; i <= R[b[l]]; i++) c[i] = a[i];
                sort(c + L[b[l]], c + R[b[l]] + 1);

				for (int i = b[l] + 1; i <= b[r] - 1; i++) tag[i] += v;

                for (int i = L[b[r]]; i <= r; i++) a[i] += v;
                for (int i = L[b[r]]; i <= R[b[r]]; i++) c[i] = a[i];
                sort(c + L[b[r]], c + R[b[r]] + 1);
                
            }
        } else {
            int ans = 0;
            if (b[l] == b[r]) {
                for (int i = l; i <= r; i++) {
                    ans += (v * v > (a[i] + tag[b[l]]));
                }
            } else {
                // l ~ R[b[l]] 查询 < v * v - tag[b[l]]
                for (int i = l; i <= R[b[l]]; i++) {
                    ans += (v * v > (a[i] + tag[b[l]]));
                }
                
                // b[l] + 1 ~ b[r] - 1  这些块内的答案
                for (int i = b[l] + 1; i <= b[r] - 1; i++) {
                    ans += lower_bound(c + L[i], c + R[i] + 1, v * v - tag[i]) -
                           (c + L[i]);
                }
				
				// L[b[r]] ~ r 查询 < v * v - tag[b[r]]
                for (int i = L[b[r]]; i <= r; i++) {
                    ans += (v * v > (a[i] + tag[b[r]]));
                }
            }
            cout << ans << '\n';
        }
    }
    return 0;
}