Problem: There is an infinite line consisting of cells. There are n boxes in some cells of this line. The i-th box stands in the cell a_i and has weight w_i. All a_i are distinct, moreover, a_{i - 1} < a_i holds for all valid i.

You would like to put together some boxes. Putting together boxes with indices in the segment [l, r] means that you will move some of them in such a way that their positions will form some segment [x, x + (r - l)].

In one step you can move any box to a neighboring cell if it isn't occupied by another box (i.e. you can choose i and change a_i by 1, all positions should remain distinct). You spend w_i units of energy moving the box i by one cell. You can move any box any number of times, in arbitrary order.

Sometimes weights of some boxes change, so you have queries of two types: 

  1. id nw — weight w_{id} of the box id becomes nw. 
  2. l r — you should compute the minimum total energy needed to put together boxes with indices in [l, r]. Since the answer can be rather big, print the remainder it gives when divided by 1000 000 007 = 10^9 + 7. Note that the boxes are not moved during the query, you only should compute the answer. 



Note that you should minimize the answer, not its remainder modulo 10^9 + 7. So if you have two possible answers 2 ⋅ 10^9 + 13 and 2 ⋅ 10^9 + 14, you should choose the first one and print 10^9 + 6, even though the remainder of the second answer is 0.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 2 ⋅ 10^5) — the number of boxes and the number of queries.

The second line contains n integers a_1, a_2, ... a_n (1 ≤ a_i ≤ 10^9) — the positions of the boxes. All a_i are distinct, a_{i - 1} < a_i holds for all valid i.

The third line contains n integers w_1, w_2, ... w_n (1 ≤ w_i ≤ 10^9) — the initial weights of the boxes.

Next q lines describe queries, one query per line.

Each query is described in a single line, containing two integers x and y. If x < 0, then this query is of the first type, where id = -x, nw = y (1 ≤ id ≤ n, 1 ≤ nw ≤ 10^9). If x > 0, then the query is of the second type, where l = x and r = y (1 ≤ l_j ≤ r_j ≤ n). x can not be equal to 0.

Output

For each query of the second type print the answer on a separate line. Since answer can be large, print the remainder it gives when divided by 1000 000 007 = 10^9 + 7.

Example

Input

5 8
1 2 6 7 10
1 1 1 1 2
1 1
1 5
1 3
3 5
-3 5
-1 10
1 4
2 5


Output

0
10
3
4
18
7

Note

Let's go through queries of the example: 

  1. 1\ 1 — there is only one box so we don't need to move anything. 
  2. 1\ 5 — we can move boxes to segment [4, 8]: 1 ⋅ |1 - 4| + 1 ⋅ |2 - 5| + 1 ⋅ |6 - 6| + 1 ⋅ |7 - 7| + 2 ⋅ |10 - 8| = 10. 
  3. 1\ 3 — we can move boxes to segment [1, 3]. 
  4. 3\ 5 — we can move boxes to segment [7, 9]. 
  5. -3\ 5 — w_3 is changed from 1 to 5. 
  6. -1\ 10 — w_1 is changed from 1 to 10. The weights are now equal to w = [10, 1, 5, 1, 2]. 
  7. 1\ 4 — we can move boxes to segment [1, 4]. 
  8. 2\ 5 — we can move boxes to segment [5, 8]. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
const int INF = 1000000001;
int n, q;
long long segSum[800005], segMult[800005], w[200005], a[200005];
void update(long long seg[], int v, int l, int r, int index, long long x);
long long get(long long seg[], int v, int l, int r, int cl, int cr);
pair<int, long long> find(long long seg[], int v, int l, int r, int cl, int cr,
                          long long s);
long long norm(long long a) {
  a %= mod;
  if (a > mod) a -= mod;
  if (a < 0) a += mod;
  return a;
}
long long mult(long long a, long long b) { return norm(norm(a) * norm(b)); }
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] = norm(a[i] - i);
  }
  for (int i = 1; i <= n; i++) cin >> w[i];
  for (int i = 1; i <= n; i++) {
    update(segSum, 1, 1, n, i, w[i]);
    update(segMult, 1, 1, n, i, mult(w[i], a[i]));
  }
  while (q--) {
    int x, y;
    cin >> x >> y;
    if (x < 0) {
      x *= -1;
      w[x] = y;
      update(segSum, 1, 1, n, x, w[x]);
      update(segMult, 1, 1, n, x, mult(w[x], a[x]));
    } else {
      long long result = 0;
      auto pos = find(segSum, 1, 1, n, x, y, get(segSum, 1, 1, n, x, y) / 2);
      assert((pos.first >= x) && (pos.first <= y));
      result = norm(mult(a[pos.first], get(segSum, 1, 1, n, x, pos.first)) -
                    norm(get(segMult, 1, 1, n, x, pos.first)));
      result = norm(result + norm(get(segMult, 1, 1, n, pos.first, y)) -
                    mult(a[pos.first], get(segSum, 1, 1, n, pos.first, y)));
      cout << result << "\n";
    }
  }
  cout << "\n"
       << "\n";
  return 0;
}
void update(long long seg[], int v, int l, int r, int index, long long x) {
  if (l == r) {
    seg[v] = x;
    return;
  }
  int mid = (l + r) >> 1;
  if (index <= mid)
    update(seg, v * 2, l, mid, index, x);
  else
    update(seg, v * 2 + 1, mid + 1, r, index, x);
  seg[v] = seg[v * 2] + seg[v * 2 + 1];
}
long long get(long long seg[], int v, int l, int r, int cl, int cr) {
  if ((l == cl) && (r == cr)) return seg[v];
  int mid = (l + r) >> 1;
  long long result = 0;
  if (cl <= mid) result = get(seg, v * 2, l, mid, cl, min(cr, mid));
  if (cr > mid) result += get(seg, v * 2 + 1, mid + 1, r, max(cl, mid + 1), cr);
  return result;
}
pair<int, long long> find(long long seg[], int v, int l, int r, int cl, int cr,
                          long long s) {
  if ((l == cl) && (r == cr)) {
    if (seg[v] <= s) return {INF, seg[v]};
    if (l == r) return {l, 0};
  }
  int mid = (l + r) >> 1;
  pair<int, long long> result = {INF, 0};
  if (cl <= mid) result = find(seg, v * 2, l, mid, cl, min(mid, cr), s);
  if ((cr > mid) && (result.first == INF)) {
    auto res = find(seg, v * 2 + 1, mid + 1, r, max(mid + 1, cl), cr,
                    s - result.second);
    result.first = res.first;
    result.second += res.second;
  }
  return result;
}