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 ppp {
  long long s, sum, res;
} t[220000 * 4], ans;
int P[220000], re[220000][3], i, j, k, l, n, m, x, y, tot, a[220000], h[220000];
int Find(int x) {
  int l = 1, r = tot, mid;
  while (l <= r) {
    mid = (l + r) / 2;
    if (P[mid] <= x)
      l = mid + 1;
    else
      r = mid - 1;
  }
  return l;
}
void renew(ppp &a, const ppp &b, const ppp &c) {
  a.sum = b.sum + c.sum;
  a.s = b.s + c.s;
  a.res = b.res + c.res + c.sum * b.s - b.sum * c.s;
}
void add(int p, int l, int r, int x, int k) {
  int mid = (l + r) / 2;
  if (l == r) {
    t[p].sum += k * P[x];
    t[p].s += k;
    return;
  }
  if (x <= mid)
    add(p * 2, l, mid, x, k);
  else
    add(p * 2 + 1, mid + 1, r, x, k);
  renew(t[p], t[p * 2], t[p * 2 + 1]);
}
void calc(int p, int l, int r, int x, int y) {
  if (x > y) return;
  int mid = (l + r) / 2;
  if (x <= l && y >= r) {
    renew(ans, ans, t[p]);
    return;
  }
  if (x <= mid) calc(p * 2, l, mid, x, y);
  if (y > mid) calc(p * 2 + 1, mid + 1, r, x, y);
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) scanf("%d", &a[i]), P[++tot] = a[i], h[i] = a[i];
  scanf("%d", &m);
  for (i = 1; i <= m; ++i) {
    for (j = 0; j < 3; ++j) scanf("%d", &re[i][j]);
    if (re[i][0] == 1) a[re[i][1]] += re[i][2], P[++tot] = a[re[i][1]];
  }
  sort(P + 1, P + tot + 1);
  for (i = 1; i <= n; ++i) add(1, 1, tot, Find(h[i]) - 1, 1);
  for (i = 1; i <= m; ++i) {
    x = re[i][1];
    y = re[i][2];
    if (re[i][0] == 1) {
      add(1, 1, tot, Find(h[x]) - 1, -1);
      h[x] += y;
      add(1, 1, tot, Find(h[x]) - 1, 1);
    } else {
      ans.s = ans.sum = ans.res = 0;
      calc(1, 1, tot, Find(x - 1), Find(y) - 1);
      printf("%I64d\n", ans.res);
    }
  }
}
