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 N = 1e5 + 5;
long long n, m, a[N], type, l, r;
inline long long myrand() { return rand() ^ (rand() << 16); }
struct item {
  item *l, *r;
  long long sum, ans;
  long long cnt, prior, key;
  item() {}
  item(long long _key) {
    key = _key;
    sum = _key;
    ans = 0;
    prior = myrand();
    cnt = 1;
    l = r = nullptr;
  }
};
typedef item *pitem;
pitem t;
inline long long cnt(pitem v) { return v ? v->cnt : 0; }
inline long long sum(pitem v) { return v ? v->sum : 0; }
inline long long ans(pitem v) { return v ? v->ans : 0; }
inline void upd_cnt(pitem &v) {
  if (v) {
    v->cnt = cnt(v->l) + 1 + cnt(v->r);
    v->sum = sum(v->l) + v->key + sum(v->r);
    v->ans =
        ans(v->l) + ans(v->r) + sum(v->r) * cnt(v->l) - sum(v->l) * cnt(v->r);
    v->ans += cnt(v->l) * v->key - sum(v->l) + sum(v->r) - cnt(v->r) * v->key;
  }
}
pitem aa(pitem l, pitem r) {
  if (!l || !r) {
    return l ? l : r;
  }
  if (l->prior > r->prior) {
    l->r = aa(l->r, r);
    upd_cnt(l);
    return l;
  } else {
    r->l = aa(l, r->l);
    upd_cnt(r);
    return r;
  }
}
void ab(pitem v, long long key, pitem &l, pitem &r) {
  if (!v) {
    l = r = nullptr;
    return;
  }
  if (key < v->key) {
    ab(v->l, key, l, v->l);
    r = v;
    upd_cnt(r);
  } else {
    ab(v->r, key, v->r, r);
    l = v;
    upd_cnt(l);
  }
}
inline void ac(long long x) {
  pitem r;
  ab(t, x - 1, t, r);
  t = aa(t, aa(new item(x), r));
}
inline void ad(long long x) {
  pitem l, r;
  ab(t, x - 1, t, r);
  ab(r, x, l, r);
  t = aa(t, r);
}
void output(pitem v) {
  if (!v) {
    return;
  }
  output(v->l);
  cout << v->key << " ";
  output(v->r);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  t = nullptr;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    ac(a[i]);
  }
  cin >> m;
  for (long long i = 1; i <= m; i++) {
    cin >> type >> l >> r;
    if (type == 1) {
      ad(a[l]);
      a[l] += r;
      ac(a[l]);
    } else {
      pitem tl, tr;
      ab(t, l - 1, t, tl);
      ab(tl, r, tl, tr);
      cout << (!tl ? 0 : tl->ans) << "\n";
      t = aa(t, aa(tl, tr));
    }
  }
}
