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>
const int oo = 1000000005, N = 5000005;
struct note {
  int c;
  long long s, f;
} T[N];
long long a[N];
int L[N], R[N], tot = 1;
note update(note L, note R) {
  note T;
  T.c = L.c + R.c;
  T.s = L.s + R.s;
  T.f = L.f + R.f + R.s * L.c - L.s * R.c;
  return T;
}
void modify(int &t, int l, int r, int x, int v) {
  if (!t) t = ++tot;
  if (l == r) {
    T[t].c += v, T[t].s += (x - oo) * v;
    return;
  }
  int mid = (0LL + l + r) / 2;
  if (x <= mid)
    modify(L[t], l, mid, x, v);
  else
    modify(R[t], mid + 1, r, x, v);
  T[t] = update(T[L[t]], T[R[t]]);
}
note query(int t, int l, int r, int ll, int rr) {
  if (!t) return T[0];
  if (l >= ll && r <= rr) return T[t];
  int mid = (0LL + l + r) / 2;
  if (rr <= mid) return query(L[t], l, mid, ll, rr);
  if (ll > mid) return query(R[t], mid + 1, r, ll, rr);
  return update(query(L[t], l, mid, ll, rr), query(R[t], mid + 1, r, ll, rr));
}
int main() {
  int n, m, x, y, z, R = 1;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x), a[i] = x + oo;
    if (a[i] >= 0 && a[i] <= 2 * oo) modify(R, 0, oo * 2, a[i], 1);
  }
  scanf("%d", &m);
  while (m--) {
    scanf("%d%d%d", &z, &x, &y);
    if (z == 1) {
      if (a[x] >= 0 && a[x] <= 2 * oo) modify(R, 0, oo * 2, a[x], -1);
      a[x] += y;
      if (a[x] >= 0 && a[x] <= 2 * oo) modify(R, 0, oo * 2, a[x], 1);
    } else
      printf("%I64d\n", query(R, 0, oo * 2, x + oo, y + oo).f);
  }
  return 0;
}
