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, m, a[200010], p[200010], L, p0[200010], p1[200010], p2[200010];
long long f[200010 * 4], g[200010 * 4], h[200010 * 4];
void add(int p, int l, int r, int x, int y, int z) {
  if (l == r - 1) {
    g[p] += y, f[p] += 0, h[p] += z;
    return;
  }
  int mid = (l + r) / 2;
  if (x < mid)
    add(p * 2 + 1, l, mid, x, y, z);
  else
    add(p * 2 + 2, mid, r, x, y, z);
  f[p] = f[p * 2 + 1] + f[p * 2 + 2] + h[p * 2 + 1] * g[p * 2 + 2];
  g[p] = g[p * 2 + 1] + g[p * 2 + 2];
  h[p] = h[p * 2 + 1] + h[p * 2 + 2];
}
pair<pair<long long, long long>, long long> ask(int p, int l, int r, int L,
                                                int R) {
  if (L <= l && r <= R) return make_pair(make_pair(f[p], g[p]), h[p]);
  int mid = (l + r) / 2;
  if (mid <= L) return ask(p * 2 + 2, mid, r, L, R);
  if (R <= mid) return ask(p * 2 + 1, l, mid, L, R);
  pair<pair<long long, long long>, long long> pl = ask(p * 2 + 1, l, mid, L, R),
                                              pr = ask(p * 2 + 2, mid, r, L, R);
  return make_pair(
      make_pair(pl.first.first + pr.first.first + pr.first.second * pl.second,
                pl.first.second + pr.first.second),
      pl.second + pr.second);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", a + i), p[L++] = a[i];
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", p0 + i, p1 + i, p2 + i);
    if (p0[i] == 1) a[p1[i] - 1] += p2[i], p[L++] = a[p1[i] - 1];
  }
  for (int i = m - 1; i >= 0; i--)
    if (p0[i] == 1) a[p1[i] - 1] -= p2[i];
  sort(p, p + L), L = unique(p, p + L) - p;
  for (int i = 0; i < n; i++)
    add(0, 0, L, lower_bound(p, p + L, a[i]) - p, a[i], 1);
  for (int i = 0; i < m; i++) {
    int x = p1[i], y = p2[i];
    if (p0[i] == 1) {
      x--;
      add(0, 0, L, lower_bound(p, p + L, a[x]) - p, -a[x], -1);
      a[x] += y;
      add(0, 0, L, lower_bound(p, p + L, a[x]) - p, a[x], 1);
    } else {
      int pl = lower_bound(p, p + L, x) - p,
          pr = lower_bound(p, p + L, y + 1) - p;
      if (pl >= pr) {
        puts("0");
        continue;
      }
      pair<pair<long long, long long>, long long> S = ask(0, 0, L, pl, pr);
      printf("%I64d\n", S.first.first * 2 - (S.second - 1) * S.first.second);
    }
  }
  return 0;
}
