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;
int n, nQ;
vector<int> x;
struct Query {
  int t, a1, a2;
  void read() { scanf("%d%d%d", &t, &a1, &a2); }
};
vector<Query> qs;
vector<int> cx;
vector<int> allx;
vector<bool> has;
struct Data {
  bool exist;
  int cnt;
  long long sum, op;
  Data() {
    sum = op = cnt = 0;
    exist = false;
  }
  Data(int v) {
    sum = v;
    op = 0;
    cnt = 1;
    exist = true;
  }
};
Data operator+(const Data& a, const Data& b) {
  if (!a.exist) return b;
  if (!b.exist) return a;
  Data ret;
  ret.exist = true;
  ret.sum = a.sum + b.sum;
  ret.op = a.op + b.op - b.cnt * a.sum + a.cnt * b.sum;
  ret.cnt = a.cnt + b.cnt;
  return ret;
}
struct Tree {
  Data data;
  Tree *pl, *pr;
  int l, r;
  void update() { data = pl->data + pr->data; }
  Tree(int l, int r) : l(l), r(r) {
    if (l + 1 == r) {
      if (has[l]) {
        data = Data(allx[l]);
      } else {
        data = Data();
      }
      return;
    }
    pl = new Tree(l, (l + r) >> 1);
    pr = new Tree((l + r) >> 1, r);
    update();
  }
  Data calc(int L, int R) {
    if (L <= l && R >= r) return data;
    if (L >= r || l >= R) return Data();
    return pl->calc(L, R) + pr->calc(L, R);
  }
  void change(int i) {
    if (i < l || i >= r) return;
    if (l + 1 == r) {
      if (has[i]) {
        has[i] = false;
        data = Data();
      } else {
        data = Data(allx[i]);
        has[i] = true;
      }
      return;
    }
    pl->change(i);
    pr->change(i);
    update();
    return;
  }
} * rt;
int get(int x) {
  return lower_bound(allx.begin(), allx.end(), x) - allx.begin();
}
int main() {
  cin >> n;
  x.assign(n, 0);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &x[i]);
  }
  allx = x;
  cx = x;
  cin >> nQ;
  qs.resize(nQ);
  for (int i = 0; i < nQ; ++i) {
    qs[i].read();
  }
  for (int i = 0; i < nQ; ++i) {
    if (qs[i].t == 1) {
      int p = qs[i].a1, d = qs[i].a2;
      --p;
      cx[p] += d;
      allx.push_back(cx[p]);
    }
  }
  sort(allx.begin(), allx.end());
  allx.erase(unique(allx.begin(), allx.end()), allx.end());
  has.assign(allx.size(), false);
  for (int i = 0; i < n; ++i) {
    has[get(x[i])] = true;
  }
  rt = new Tree(0, allx.size());
  cx = x;
  for (int i = 0; i < qs.size(); ++i) {
    Query q = qs[i];
    if (q.t == 1) {
      int p = q.a1, d = q.a2;
      --p;
      int x = cx[p];
      int nx = cx[p] += d;
      rt->change(get(x));
      rt->change(get(nx));
    } else {
      int l = q.a1, r = q.a2;
      l = lower_bound(allx.begin(), allx.end(), l) - allx.begin();
      r = upper_bound(allx.begin(), allx.end(), r) - allx.begin();
      Data ret = rt->calc(l, r);
      cout << ret.op << endl;
    }
  }
  return 0;
}
