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 MAX = 1e5 + 9, INF = 1200000000;
int pos[MAX], n, q, type;
long long l, r;
struct node {
  int l, r;
  long long ldis, rdis, dis, cnt;
  node *le, *ri;
  node(long long _l = -INF, long long _r = INF) {
    l = _l, r = _r, le = ri = NULL, cnt = ldis = rdis = dis = 0;
  }
} root;
node merge(node &left, node &right) {
  node ret(left.l, right.r);
  ret.le = &left, ret.ri = &right;
  ret.cnt = left.cnt + right.cnt;
  ret.ldis = left.ldis + right.ldis + (left.r - left.l) * right.cnt,
  ret.rdis = right.rdis + left.rdis + (right.r - right.l) * left.cnt;
  ret.dis =
      left.dis + right.dis + left.rdis * right.cnt + right.ldis * left.cnt;
  return ret;
}
void add(long long pos, long long val, node &now) {
  if (now.r - now.l == 1) {
    now.cnt += val, now.rdis += val;
    return;
  }
  long long mid = now.l + now.r >> 1;
  if (now.le == NULL) now.le = new node(now.l, mid);
  if (now.ri == NULL) now.ri = new node(mid, now.r);
  if (pos < mid)
    add(pos, val, *now.le);
  else
    add(pos, val, *now.ri);
  now = merge(*now.le, *now.ri);
  return;
}
node get(long long s, long long e, node &now) {
  if (s <= now.l && now.r <= e) return now;
  long long mid = now.l + now.r >> 1;
  if (s < mid && now.le == NULL) now.le = new node(now.l, mid);
  if (e > mid && now.ri == NULL) now.ri = new node(mid, now.r);
  if (e <= mid) return get(s, e, *now.le);
  if (s >= mid) return get(s, e, *now.ri);
  node ret1, ret2;
  return merge(ret1 = get(s, e, *now.le), (ret2 = get(s, e, *now.ri)));
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> pos[i], add(pos[i], +1, root);
  cin >> q;
  while (q--) {
    cin >> type;
    if (type == 1)
      cin >> l >> r, l--, add(pos[l], -1, root), pos[l] += r,
          add(pos[l], +1, root);
    else
      cin >> l >> r, r++, cout << get(l, r, root).dis << '\n';
  }
}
