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 int N = 1e5 + 10;
int n, m, x[N];
int ptr = 1;
struct node {
  long long sum = 0, res = 0;
  int cnt = 0, lc = 0, rc = 0;
  void make_child() {
    if (lc == 0) lc = ptr++;
    if (rc == 0) rc = ptr++;
  }
} seg[N * 50];
node merge(node a, node b) {
  node ret;
  ret.res = a.res + b.res;
  ret.res += 1LL * b.sum * a.cnt - 1LL * a.sum * b.cnt;
  ret.cnt = a.cnt + b.cnt;
  ret.sum = a.sum + b.sum;
  return ret;
}
void add(int p, int val, int id = 0, long long l = -2e9, long long r = 2e9) {
  if (r - l == 1) {
    seg[id].res = 0;
    seg[id].cnt += val;
    seg[id].sum += val * p;
    return;
  }
  seg[id].make_child();
  int lc = seg[id].lc, rc = seg[id].rc;
  long long mid = (l + r) / 2;
  if (p < mid)
    add(p, val, lc, l, mid);
  else
    add(p, val, rc, mid, r);
  seg[id] = merge(seg[seg[id].lc], seg[seg[id].rc]);
  seg[id].lc = lc, seg[id].rc = rc;
}
node get(int x, int y, int id = 0, long long l = -2e9, long long r = 2e9) {
  node ret;
  if (x <= l && r <= y) return seg[id];
  if (y <= l || r <= x) return ret;
  seg[id].make_child();
  long long mid = (l + r) / 2;
  ret = merge(get(x, y, seg[id].lc, l, mid), get(x, y, seg[id].rc, mid, r));
  return ret;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x[i];
    add(x[i], 1);
  }
  cin >> m;
  while (m--) {
    int typ;
    cin >> typ;
    if (typ == 1) {
      int p, d;
      cin >> p >> d;
      p--;
      add(x[p], -1);
      x[p] += d;
      add(x[p], 1);
    } else {
      int l, r;
      cin >> l >> r;
      r++;
      cout << get(l, r).res << "\n";
    }
  }
  return 0;
}
