Yaroslav has n points that lie on the Ox axis. The coordinate of the first point is x1, the coordinate of the second point is x2, ..., the coordinate of the n-th point is — xn. Now Yaroslav wants to execute m queries, each of them is of one of the two following types:

  1. Move the pj-th point from position xpj to position xpj + dj. At that, it is guaranteed that after executing such query all coordinates of the points will be distinct. 
  2. Count the sum of distances between all pairs of points that lie on the segment [lj, rj] (lj ≤ rj). In other words, you should count the sum of: <image>. 



Help Yaroslav.

Input

The first line contains integer n — the number of points (1 ≤ n ≤ 105). The second line contains distinct integers x1, x2, ..., xn — the coordinates of points (|xi| ≤ 109).

The third line contains integer m — the number of queries (1 ≤ m ≤ 105). The next m lines contain the queries. The j-th line first contains integer tj (1 ≤ tj ≤ 2) — the query type. If tj = 1, then it is followed by two integers pj and dj (1 ≤ pj ≤ n, |dj| ≤ 1000). If tj = 2, then it is followed by two integers lj and rj ( - 109 ≤ lj ≤ rj ≤ 109).

It is guaranteed that at any moment all the points have distinct coordinates.

Output

For each type 2 query print the answer on a single line. Print the answers in the order, in which the queries follow in the input.

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.

Examples

Input

8
36 50 28 -75 40 -60 -95 -48
20
2 -61 29
1 5 -53
1 1 429
1 5 130
2 -101 -71
2 -69 53
1 1 404
1 5 518
2 -101 53
2 50 872
1 1 -207
2 -99 -40
1 7 -389
1 6 -171
1 2 464
1 7 -707
1 1 -730
1 1 560
2 635 644
1 7 -677


Output

176
20
406
1046
1638
156
0
#include <bits/stdc++.h>
using namespace std;
const long long maxn = 2e6 + 100;
const long long mod = 1e9 + 7;
const long long inf = 1e18;
struct dat {
  long long cnt, sum, dis;
};
long long n, m, k, a[maxn], b[maxn];
dat seg[maxn];
vector<long long> p;
vector<pair<long long, pair<long long, long long> > > q;
vector<dat> vec;
dat mrg(dat p1, dat p2) {
  dat p3;
  p3.cnt = p1.cnt + p2.cnt;
  p3.sum = p1.sum + p2.sum;
  p3.dis = p1.dis + p2.dis + p1.cnt * p2.sum - p1.sum * p2.cnt;
  return p3;
}
void upd(long long id, long long s, long long e, long long l, long long x,
         long long y) {
  if (l < s || e <= l) return;
  if (e - s == 1) {
    seg[id].cnt += x;
    seg[id].sum += y;
    return;
  }
  long long mid = (s + e) / 2;
  upd(id * 2, s, mid, l, x, y);
  upd(id * 2 + 1, mid, e, l, x, y);
  seg[id] = mrg(seg[id * 2], seg[id * 2 + 1]);
  return;
}
void get(long long id, long long s, long long e, long long l, long long r) {
  if (r <= s || e <= l) return;
  if (l <= s && e <= r) {
    vec.push_back(seg[id]);
    return;
  }
  long long mid = (s + e) / 2;
  get(id * 2, s, mid, l, r);
  get(id * 2 + 1, mid, e, l, r);
  return;
}
long long solve(long long l, long long r) {
  vec.clear();
  get(1, 0, k, l, r);
  for (long long i = 1; i < vec.size(); i++) {
    vec[i] = mrg(vec[i - 1], vec[i]);
  }
  if (vec.empty()) return 0;
  return vec[vec.size() - 1].dis;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t, l, r;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
    p.push_back(b[i]);
  }
  cin >> m;
  for (long long i = 0; i < m; i++) {
    cin >> t >> l >> r;
    q.push_back(make_pair(t, make_pair(l, r)));
    if (t == 1) {
      b[l - 1] += r;
      p.push_back(b[l - 1]);
    }
  }
  sort(p.begin(), p.end());
  k = p.size();
  for (long long i = 0; i < n; i++) {
    upd(1, 0, k, lower_bound(p.begin(), p.end(), a[i]) - p.begin(), 1, a[i]);
  }
  for (long long i = 0; i < m; i++) {
    t = q[i].first;
    l = q[i].second.first;
    r = q[i].second.second;
    if (t == 1) {
      upd(1, 0, k, lower_bound(p.begin(), p.end(), a[l - 1]) - p.begin(), -1,
          -a[l - 1]);
      a[l - 1] += r;
      upd(1, 0, k, lower_bound(p.begin(), p.end(), a[l - 1]) - p.begin(), 1,
          a[l - 1]);
    } else {
      cout << solve(lower_bound(p.begin(), p.end(), l) - p.begin(),
                    upper_bound(p.begin(), p.end(), r) - p.begin())
           << '\n';
    }
  }
  return 0;
}
