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 N = 1e5 + 20;
long long n, m, n_, x[N], last[N], type[N], a[N], b[N], ads[N << 1];
vector<pair<long long, long long> > vec;
struct node {
  long long sz, sum, ans;
  node() { sz = sum = ans = 0; }
} seg[N << 3];
node merge(node x, node y) {
  node ret;
  ret.sz = x.sz + y.sz;
  ret.sum = x.sum + y.sum;
  ret.ans = x.ans + y.ans + x.sz * y.sum - y.sz * x.sum;
  return ret;
}
void build(long long v = 1, long long s = 0, long long e = n) {
  if (s + 1 == e) {
    if (vec[s].second < n_) {
      seg[v].sum = vec[s].first;
      seg[v].sz = 1;
    }
    return;
  }
  long long mid = s + e >> 1, lc = v << 1, rc = lc | 1;
  build(lc, s, mid);
  build(rc, mid, e);
  seg[v] = merge(seg[lc], seg[rc]);
}
node get(long long l, long long r, long long v = 1, long long s = 0,
         long long e = n) {
  if (s >= r || e <= l) {
    node ret;
    return ret;
  }
  if (s >= l && e <= r) return seg[v];
  long long mid = s + e >> 1, lc = v << 1, rc = lc | 1;
  return merge(get(l, r, lc, s, mid), get(l, r, rc, mid, e));
}
void upd(long long idx, long long v = 1, long long s = 0, long long e = n) {
  if (idx < s || idx >= e) return;
  if (s + 1 == e) {
    seg[v].sum = vec[s].first - seg[v].sum;
    seg[v].sz = 1 - seg[v].sz;
    return;
  }
  long long mid = s + e >> 1, lc = v << 1, rc = lc | 1;
  upd(idx, lc, s, mid);
  upd(idx, rc, mid, e);
  seg[v] = merge(seg[lc], seg[rc]);
  return;
}
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 >> x[i];
    last[i] = i;
    vec.push_back({x[i], i});
  }
  cin >> m;
  for (long long i = 0; i < m; i++) {
    cin >> type[i] >> a[i] >> b[i];
    if (type[i] == 1) {
      a[i]--;
      x[a[i]] += b[i];
      vec.push_back({x[a[i]], n + i});
    }
  }
  sort(vec.begin(), vec.end());
  n_ = n, n = vec.size();
  build();
  for (long long i = 0; i < n; i++) ads[vec[i].second] = i;
  for (long long i = 0; i < m; i++) {
    if (type[i] == 1) {
      upd(ads[last[a[i]]]);
      upd(ads[n_ + i]);
      last[a[i]] = n_ + i;
    } else {
      long long l = lower_bound(vec.begin(), vec.end(),
                                pair<long long, long long>(a[i], -1)) -
                    vec.begin();
      long long r = lower_bound(vec.begin(), vec.end(),
                                pair<long long, long long>(b[i] + 1, -1)) -
                    vec.begin();
      cout << get(l, r).ans << '\n';
    }
  }
  return 0;
}
