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;
struct Node {
  long long cnt, sum, ans;
  int ls, rs;
  Node() { cnt = sum = ans = ls = rs = 0; }
} c[5000005], zero;
int n, m, x[500005], vvv = 1;
Node merge(Node A, Node B, Node C) {
  C.cnt = A.cnt + B.cnt;
  C.sum = A.sum + B.sum;
  C.ans = A.cnt * B.sum - B.cnt * A.sum + A.ans + B.ans;
  return C;
}
void add(int v, int z, int p, int l, int r) {
  if (l > v || r < v) return;
  if (l == r) {
    c[p].sum += v * z, c[p].cnt += z, c[p].ans = 0;
    return;
  }
  if (!c[p].ls) c[p].ls = ++vvv;
  if (!c[p].rs) c[p].rs = ++vvv;
  int mid = l + r >> 1;
  if (mid >= v) add(v, z, c[p].ls, l, mid);
  if (mid < v) add(v, z, c[p].rs, mid + 1, r);
  c[p] = merge(c[c[p].ls], c[c[p].rs], c[p]);
}
Node query(int x, int y, int p, int l, int r) {
  if (x <= l && r <= y) return c[p];
  int mid = l + r >> 1;
  Node A, B;
  if (mid >= x) A = query(x, y, c[p].ls, l, mid);
  if (mid < y) B = query(x, y, c[p].rs, mid + 1, r);
  return merge(A, B, c[p]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x[i]);
    add(x[i], 1, 1, -1000000001, 1000000001);
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    int od;
    scanf("%d", &od);
    if (od == 1) {
      int a, b;
      scanf("%d%d", &a, &b);
      add(x[a], -1, 1, -1000000001, 1000000001);
      x[a] += b;
      add(x[a], 1, 1, -1000000001, 1000000001);
    } else {
      int a, b;
      scanf("%d%d", &a, &b);
      printf("%lld\n", query(a, b, 1, -1000000001, 1000000001).ans);
    }
  }
  return 0;
}
