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 int maxn = 1e5 + 10;
struct node {
  int s, e;
  long long ans, l, r, cnt;
  node *bl, *br;
  node(int ss = -1.1e9 - 10, int ee = 1.1e9 + 10) {
    s = ss, e = ee, l = r = cnt = ans = 0, bl = br = NULL;
  }
} root;
int n, q, a[maxn];
node merge(node &le, node &ri) {
  node ret(le.s, ri.e);
  ret.bl = &le, ret.br = &ri;
  ret.ans = le.ans + ri.ans + le.r * ri.cnt + ri.l * le.cnt;
  ret.l = le.l + ri.l + ri.cnt * (le.e - le.s);
  ret.r = le.r + ri.r + le.cnt * (ri.e - ri.s);
  ret.cnt = le.cnt + ri.cnt;
  return ret;
}
void upd(long long x, long long val, node &now) {
  if (now.e == now.s + 1) {
    now.r += val, now.cnt += val;
    return;
  }
  long long mid = now.s + now.e >> 1;
  if (now.bl == NULL) now.bl = new node(now.s, mid);
  if (now.br == NULL) now.br = new node(mid, now.e);
  if (x < mid)
    upd(x, val, *now.bl);
  else
    upd(x, val, *now.br);
  now = merge(*now.bl, *now.br);
}
node get(long long l, long long r, node &now) {
  if (l <= now.s && now.e <= r) return now;
  long long mid = now.s + now.e >> 1;
  if (l < mid && now.bl == NULL) now.bl = new node(now.s, mid);
  if (r > mid && now.br == NULL) now.br = new node(mid, now.e);
  if (r <= mid) return get(l, r, *now.bl);
  if (l >= mid) return get(l, r, *now.br);
  node ret1, ret2;
  return merge(ret1 = get(l, r, *now.bl), ret2 = get(l, r, *now.br));
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i], upd(a[i], 1, root);
  cin >> q;
  while (q--) {
    long long t, l, r;
    cin >> t >> l >> r;
    if (t == 1) l--, upd(a[l], -1, root), a[l] += r, upd(a[l], 1, root);
    if (t == 2) cout << get(l, r + 1, root).ans << "\n";
  }
}
