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;
long long pos[100005];
long long n;
struct Node {
  Node *L, *R;
  long long sum, cnt, ans;
  Node() {
    L = R = NULL;
    sum = cnt = ans = 0;
  }
} root, zero;
void make(Node &id) {
  if (id.L == NULL) id.L = new Node();
  if (id.R == NULL) id.R = new Node();
  return;
}
Node merge(Node c, Node a, Node b) {
  c.sum = a.sum + b.sum;
  c.cnt = a.cnt + b.cnt;
  c.ans = a.ans + b.ans + a.cnt * b.sum - b.cnt * a.sum;
  return c;
}
void add(long long p, long long val, long long l = -2000000100,
         long long r = 2000000100, Node &id = root) {
  if (p < l || p >= r) return;
  if (r - l == 1) {
    id.ans = 0;
    id.cnt += val;
    id.sum += p * val;
    return;
  }
  make(id);
  long long mid = (l + r) / 2;
  add(p, val, l, mid, *id.L);
  add(p, val, mid, r, *id.R);
  id = merge(id, *id.L, *id.R);
}
Node get(long long x, long long y, long long l = -2000000100,
         long long r = 2000000100, Node &id = root) {
  if (y <= l || x >= r) return zero;
  if (x <= l && y >= r) return id;
  long long mid = (l + r) / 2;
  Node rl, rr;
  if (id.L != NULL) rl = get(x, y, l, mid, *id.L);
  if (id.R != NULL) rr = get(x, y, mid, r, *id.R);
  return merge(id, rl, rr);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long p;
    cin >> p;
    pos[i] = p;
    add(pos[i], 1);
  }
  long long q;
  cin >> q;
  while (q--) {
    long long type;
    cin >> type;
    if (type == 2) {
      long long x, y;
      cin >> x >> y;
      cout << get(x, y + 1).ans << endl;
    } else {
      long long i, p;
      cin >> i >> p;
      i--;
      add(pos[i], -1);
      pos[i] += p;
      add(pos[i], 1);
    }
  }
}
