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;
long long const MN = 1e5 + 100, INF = 2e9 + 100;
long long n, q, lst = 2, a[MN];
struct node {
  node *L, *R;
  long long sum, cnt, ans;
  node() {
    L = R = NULL;
    ans = cnt = sum = 0;
  }
} root, zero;
void make(node &a) {
  if (a.L == NULL) a.L = new node();
  if (a.R == NULL) a.R = new node();
}
node merge(node ret, node a, node b) {
  ret.sum = a.sum + b.sum;
  ret.cnt = a.cnt + b.cnt;
  ret.ans = a.ans + b.ans + a.cnt * b.sum - b.cnt * a.sum;
  return ret;
}
void upd(long long ind, long long k, long long l = -INF, long long r = INF,
         node &C = root) {
  if (ind < l || ind >= r) return;
  if (r - l == 1) {
    C.ans = 0;
    C.cnt += k;
    C.sum += k * ind;
    return;
  }
  make(C);
  long long m = (l + r) / 2;
  upd(ind, k, l, m, *C.L), upd(ind, k, m, r, *C.R);
  C = merge(C, *C.L, *C.R);
}
node get(long long s, long long e, long long l = -INF, long long r = INF,
         node &C = root) {
  if (l >= e || r <= s) return zero;
  if (l >= s && r <= e) return C;
  long long m = (l + r) / 2;
  node rl, rr;
  if (C.L != NULL) rl = get(s, e, l, m, *C.L);
  if (C.R != NULL) rr = get(s, e, m, r, *C.R);
  return merge(C, rl, rr);
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    upd(a[i], 1);
  }
  cin >> q;
  while (q--) {
    long long ch, l, r, i, d;
    cin >> ch;
    if (ch == 1) {
      cin >> i >> d;
      i--;
      upd(a[i], -1);
      a[i] += d;
      upd(a[i], 1);
    } else {
      cin >> l >> r;
      cout << get(l, r + 1).ans << '\n';
    }
  }
}
