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 = (long long)1e5 + 10;
const long long oo = 1e10;
long long n, m, xl[N * 30], xr[N * 30], ptr = 2;
long long a[N];
struct data {
  long long res, sum;
  long long cnt;
  data() { res = cnt = sum = 0; }
} seg[N * 30], null;
data merge(data lc, data rc) {
  if (lc.cnt == -1) return rc;
  if (rc.cnt == -1) return lc;
  data ret;
  ret.cnt = lc.cnt + rc.cnt;
  ret.sum = lc.sum + rc.sum;
  long long add = rc.sum * lc.cnt - lc.sum * rc.cnt;
  ret.res = lc.res + rc.res + add;
  return ret;
}
void rebuild(long long pos, long long ch, long long id = 1, long long l = 0,
             long long r = 2 * oo) {
  if (r - l < 2) {
    if (~ch)
      ++seg[id].cnt, seg[id].sum = pos;
    else
      --seg[id].cnt, seg[id].sum = 0;
    return;
  }
  long long mid = r + l >> 1;
  if (pos < mid) {
    if (!xl[id]) xl[id] = ptr++;
    rebuild(pos, ch, xl[id], l, mid);
  } else {
    if (!xr[id]) xr[id] = ptr++;
    rebuild(pos, ch, xr[id], mid, r);
  }
  if (xl[id] && xr[id])
    seg[id] = merge(seg[xl[id]], seg[xr[id]]);
  else if (xl[id])
    seg[id] = seg[xl[id]];
  else
    seg[id] = seg[xr[id]];
}
data get(long long x, long long y, long long id = 1, long long l = 0,
         long long r = 2 * oo) {
  if (r <= x || l >= y) return null;
  if (l >= x && r <= y) return seg[id];
  long long mid = r + l >> 1;
  data lc, rc;
  lc.cnt = rc.cnt = -1;
  if (xl[id]) lc = get(x, y, xl[id], l, mid);
  if (xr[id]) rc = get(x, y, xr[id], mid, r);
  return merge(lc, rc);
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  null.cnt = -1;
  cin >> n;
  for (long long j = 0; j < n; ++j) cin >> a[j], rebuild(a[j] + oo, +1);
  cin >> m;
  while (m--) {
    long long typ;
    cin >> typ;
    if (typ == 1) {
      long long p, d;
      cin >> p >> d;
      --p;
      rebuild(a[p] + oo, -1);
      a[p] += d;
      rebuild(a[p] + oo, +1);
    } else {
      long long l, r;
      cin >> l >> r;
      cout << get(l + oo, r + oo + 1).res << endl;
    }
  }
}
