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 I = 1e10, M = 1e5 + 10;
long long a[M], x, y, q, n, i, o;
struct kh {
  long long cn, S;
  long long an;
  kh *r, *l;
  long long s, e;
  kh(long long x, long long y) {
    s = x;
    e = y;
    r = l = NULL;
    cn = S = an = 0;
  }
};
kh z = *(new kh(0, 0)), *ro = new kh(-I, I);
kh m1(kh x, kh y) {
  kh ret = z;
  ret.cn = x.cn + y.cn;
  ret.S = x.S + y.S;
  ret.an = x.an + y.an;
  ret.an -= x.S * y.cn;
  ret.an += y.S * x.cn;
  return ret;
}
void ad(long long ind2, long long val, kh *p) {
  if (p->s + 1 == p->e) {
    p->cn += val;
    if (val == 1)
      p->S = ind2;
    else
      p->S = 0;
    return;
  }
  long long mid = (p->s + p->e) / 2;
  if (ind2 < mid) {
    if (p->l == NULL) p->l = new kh(p->s, mid);
    ad(ind2, val, p->l);
  }
  if (ind2 >= mid) {
    if (p->r == NULL) p->r = new kh(mid, p->e);
    ad(ind2, val, p->r);
  }
  kh t = z;
  if (p->r == NULL)
    t = *(p->l);
  else if (p->l == NULL)
    t = *(p->r);
  else
    t = m1(*(p->l), *(p->r));
  p->an = t.an;
  p->S = t.S;
  p->cn = t.cn;
}
kh fi(long long x, long long y, kh *p) {
  if (p == NULL) return z;
  long long st = p->s, en = p->e;
  if (st >= y or en <= x) return z;
  if (st >= x and en <= y) return *p;
  long long mid = (st + en) / 2;
  kh an1 = fi(x, y, p->l), an2 = fi(x, y, p->r);
  return m1(an1, an2);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> x;
    a[i] = x;
    ad(x, 1, ro);
  }
  cin >> q;
  for (i = 1; i <= q; i++) {
    cin >> o >> x >> y;
    if (o == 1) {
      ad(a[x], -1, ro);
      a[x] += y;
      ad(a[x], 1, ro);
    } else
      cout << fi(x, y + 1, ro).an << '\n';
  }
}
