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;
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long nmax = 100005;
long long n, a[nmax], m, b[nmax];
vector<long long> vecc;
vector<pair<long long, pair<long long, long long>>> Queries;
struct item {
  long long nr_el, val, sum;
  item() { nr_el = val = sum = 0; }
};
vector<item> tree(16 * nmax);
item _merge(item A, item B) {
  item ANS;
  ANS.nr_el = A.nr_el + B.nr_el;
  ANS.sum = A.sum + B.sum;
  ANS.val = A.val + B.val - A.sum * B.nr_el + B.sum * A.nr_el;
  return ANS;
}
void update(long long l, long long r, long long pos, long long val1,
            long long val2, long long nod) {
  if (l == r) {
    tree[nod].nr_el += val1;
    tree[nod].sum += val2;
  } else {
    long long mid = l + (r - l) / 2;
    if (pos <= mid) {
      update(l, mid, pos, val1, val2, 2 * nod);
    } else {
      update(mid + 1, r, pos, val1, val2, 2 * nod + 1);
    }
    tree[nod] = _merge(tree[2 * nod], tree[2 * nod + 1]);
  }
}
item query(long long l, long long r, long long le, long long ri,
           long long nod) {
  if (l > ri || r < le) {
    item lol;
    return lol;
  } else if (l >= le && r <= ri)
    return tree[nod];
  else {
    long long mid = l + (r - l) / 2;
    return _merge(query(l, mid, le, ri, 2 * nod),
                  query(mid + 1, r, le, ri, 2 * nod + 1));
  }
}
int32_t main() {
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    b[i] = a[i];
    vecc.push_back(a[i]);
  }
  cin >> m;
  for (long long i = 1; i <= m; i++) {
    long long type;
    cin >> type;
    if (type == 1) {
      long long x, y;
      cin >> x >> y;
      b[x] += y;
      vecc.push_back(b[x]);
      Queries.push_back({type, {x, y}});
    } else {
      long long x, y;
      cin >> x >> y;
      vecc.push_back(x);
      vecc.push_back(y);
      Queries.push_back({type, {x, y}});
    }
  }
  sort((vecc).begin(), (vecc).end());
  vecc.resize(unique((vecc).begin(), (vecc).end()) - vecc.begin());
  long long nn = (long long)vecc.size();
  for (long long i = 1; i <= n; i++) {
    long long pos =
        lower_bound((vecc).begin(), (vecc).end(), a[i]) - vecc.begin() + 1;
    update(1, nn, pos, 1, a[i], 1);
  }
  for (long long i = 1; i <= m; i++) {
    if (Queries[i - 1].first == 1) {
      long long pos = lower_bound((vecc).begin(), (vecc).end(),
                                  a[Queries[i - 1].second.first]) -
                      vecc.begin() + 1;
      update(1, nn, pos, -1, -a[Queries[i - 1].second.first], 1);
      a[Queries[i - 1].second.first] += Queries[i - 1].second.second;
      pos = lower_bound((vecc).begin(), (vecc).end(),
                        a[Queries[i - 1].second.first]) -
            vecc.begin() + 1;
      update(1, nn, pos, 1, a[Queries[i - 1].second.first], 1);
    } else {
      long long x = lower_bound((vecc).begin(), (vecc).end(),
                                Queries[i - 1].second.first) -
                    vecc.begin() + 1;
      long long y = lower_bound((vecc).begin(), (vecc).end(),
                                Queries[i - 1].second.second) -
                    vecc.begin() + 1;
      cout << query(1, nn, x, y, 1).val << '\n';
    }
  }
}
