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 = 6e5;
const long long INF = 3e18;
const long long MOD = 1000000007;
struct node {
  long long R, L, val, left, right;
} seg[N];
long long n, m, c[N], d[N];
node f(node a, node b) {
  long long l = a.R - a.L + 1;
  long long l1 = b.R - b.L + 1;
  node ans;
  ans.L = a.L;
  ans.R = b.R;
  ans.val =
      a.val + b.val + a.right * l1 + b.left * l + l * l1 * (d[b.L] - d[a.R]);
  ans.right = a.right + b.right + (d[b.R] - d[a.R]) * l;
  ans.left = a.left + b.left + (d[b.L] - d[a.L]) * l1;
  return ans;
}
node build(long long v, long long l, long long r) {
  if (l == r) {
    seg[v].val = 0;
    seg[v].R = l;
    seg[v].L = l;
    seg[v].right = 0;
    seg[v].left = 0;
    return seg[v];
  }
  seg[v] =
      f(build(2 * v, l, (l + r) / 2), build(2 * v + 1, (l + r) / 2 + 1, r));
  return seg[v];
}
node get(long long v, long long l, long long r) {
  long long mid = (seg[v].R + seg[v].L) / 2;
  if (l <= seg[v].L && r >= seg[v].R) return seg[v];
  if (r <= mid)
    return get(2 * v, l, r);
  else if (l > mid)
    return get(2 * v + 1, l, r);
  else
    return f(get(2 * v, l, r), get(2 * v + 1, l, r));
}
node update(long long v, long long l, long long r) {
  long long mid = (seg[v].R + seg[v].L) / 2;
  if (seg[v].L == seg[v].R) return seg[v];
  if (l > mid)
    seg[v] = f(seg[2 * v], update(2 * v + 1, l, r));
  else if (r <= mid)
    seg[v] = f(update(2 * v, l, r), seg[2 * v + 1]);
  else
    seg[v] = f(update(2 * v, l, r), update(2 * v + 1, l, r));
  return seg[v];
}
int main() {
  ifstream inf("input.txt");
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
    d[i] = c[i];
  }
  sort(d + 1, d + n + 1);
  build(1, 1, n);
  cin >> m;
  for (int i = 1; i <= m; i++) {
    long long t, x, y;
    cin >> t >> x >> y;
    if (t == 2) {
      long long x1 = lower_bound(d + 1, d + n + 1, x) - d;
      long long y1 = lower_bound(d + 1, d + n + 1, y) - d;
      if (x1 == n + 1) x1--;
      if (y1 == n + 1) y1--;
      if (d[y1] > y) y1--;
      if (y1 == 0 || y1 < x1) {
        cout << 0 << endl;
        continue;
      }
      cout << get(1, x1, y1).val << endl;
    }
    if (t == 1) {
      long long h = lower_bound(d + 1, d + n + 1, c[x]) - d;
      c[x] += y;
      long long u = lower_bound(d + 1, d + n + 1, c[x]) - d;
      if (u > h) u--;
      if (u < h) {
        long long k = d[h] + y;
        for (int j = h - 1; j >= u; j--) d[j + 1] = d[j];
        d[u] = k;
      } else {
        long long k = d[h] + y;
        for (int j = h + 1; j <= u; j++) d[j - 1] = d[j];
        d[u] = k;
      }
      update(1, min(u, h), max(u, h));
    }
  }
}
