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>
#pragma comment(linker, "/stack:200000000")
using namespace std;
const int mod = 1000000007;
const int maxn = 100010;
const int inf = 2e9;
const double eps = 1e-7;
inline int gcd(int a, int b) {
  if (a < b) swap(a, b);
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
struct item {
  long long key, prior, cnt;
  item *l, *r;
  long long ans;
  long long loc[2];
  long long sum[2];
  item() {}
  item(int key, int prior) : key(key), prior(prior), l(NULL), r(NULL), cnt(1) {
    sum[0] = sum[1] = 0;
    loc[0] = loc[1] = key;
    ans = 0;
  }
};
typedef item *pitem;
long long cnt(pitem t) { return t ? t->cnt : 0; }
long long sum(pitem t, int suf) { return t ? t->sum[suf] : 0; }
long long loc(pitem t, int ind) { return t ? t->loc[ind] : (ind ? -inf : inf); }
long long ans(pitem t) { return t ? t->ans : 0; }
void upd_item(pitem t) {
  if (t) {
    t->cnt = cnt(t->l) + cnt(t->r) + 1;
    long long sum1 =
                  t->l ? sum(t->l, 1) + (t->key - loc(t->l, 1)) * cnt(t->l) : 0,
              sum2 = t->r ? sum(t->r, 0) + (-t->key + loc(t->r, 0)) * cnt(t->r)
                          : 0;
    t->sum[0] = sum(t->l, 0) + sum2 +
                (t->l ? (t->key - loc(t->l, 0)) * (cnt(t->r) + 1) : 0);
    t->sum[1] = sum(t->r, 1) + sum1 +
                (t->r ? (-t->key + loc(t->r, 1)) * (cnt(t->l) + 1) : 0);
    t->loc[0] =
        ((loc(t->l, 0) < loc(t->r, 0)) ? (loc(t->l, 0)) : (loc(t->r, 0)));
    t->loc[0] = ((t->loc[0] < t->key) ? (t->loc[0]) : (t->key));
    t->loc[1] =
        ((loc(t->l, 1) > loc(t->r, 1)) ? (loc(t->l, 1)) : (loc(t->r, 1)));
    t->loc[1] = ((t->loc[1] > t->key) ? (t->loc[1]) : (t->key));
    t->ans =
        ans(t->l) + ans(t->r) + sum1 * (cnt(t->r) + 1) + sum2 * (cnt(t->l) + 1);
  }
}
void split(pitem t, pitem &l, pitem &r, long long key) {
  if (!t) return void(l = r = 0);
  if (key <= t->key)
    split(t->l, l, t->l, key), r = t;
  else
    split(t->r, t->r, r, key), l = t;
  upd_item(t);
}
void merge(pitem l, pitem r, pitem &t) {
  if (!l || !r)
    t = l ? l : r;
  else if (l->prior > r->prior)
    merge(l->r, r, l->r), t = l;
  else
    merge(l, r->l, r->l), t = r;
  upd_item(t);
}
int get_i(pitem t, long long ind) {
  if (cnt(t->l) == ind)
    return t->key;
  else if (cnt(t->l) > ind)
    return get_i(t->l, ind);
  else
    return get_i(t->r, ind - cnt(t->l) - 1);
}
int main() {
  int n;
  cin >> n;
  int pts[maxn];
  int p[maxn];
  for (int i = (0); i < (n); i++) {
    cin >> pts[i];
    p[i] = pts[i];
  }
  sort(pts, pts + n);
  pitem tree = new item(pts[0], rand());
  for (int i = (1); i < (n); i++) {
    pitem nt = new item(pts[i], rand());
    merge(tree, nt, tree);
  }
  int m;
  cin >> m;
  while (m--) {
    long long t, x, y;
    cin >> t >> x >> y;
    if (t == 1) {
      x--;
      long long val = p[x];
      pitem tmp1, tmp2;
      split(tree, tmp1, tree, val);
      split(tree, tree, tmp2, val + 1);
      merge(tmp1, tmp2, tree);
      val += y;
      p[x] += y;
      pitem nt = new item(val, rand());
      split(tree, tmp1, tmp2, val);
      merge(tmp1, nt, tree);
      merge(tree, tmp2, tree);
    } else {
      pitem tmp1, tmp2;
      split(tree, tmp1, tree, x);
      split(tree, tree, tmp2, y + 1);
      cout << ans(tree) << endl;
      merge(tmp1, tree, tree);
      merge(tree, tmp2, tree);
    }
  }
}
