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 long long MOD = 1e9 + 7;
const int N = 1 << 18;
long long places[N];
long long weights[N];
long long negs[2 * N];
long long mults[2 * N];
void update(int i) {
  negs[i] = (negs[2 * i] + negs[2 * i + 1]) % MOD;
  mults[i] = mults[2 * i] + mults[2 * i + 1];
}
void setWeight(int i, int w) {
  int j = i + N;
  mults[j] = w;
  negs[j] = ((places[i] - i) * w) % MOD;
  j >>= 1;
  while (j > 0) {
    update(j);
    j >>= 1;
  }
}
long long ans(int a, int b, int j, int i = 1, int ia = 0, int ib = N - 1) {
  if (b < ia || ib < a) return 0;
  if (a <= ia && ib <= b) {
    long long mt = (mults[i] % MOD) * (places[j] - j);
    long long nt = negs[i];
    if (ib <= j) {
      return (mt - nt) % MOD;
    } else if (ia >= j) {
      return (nt - mt) % MOD;
    }
  }
  int mid = (ia + ib) >> 1;
  return (ans(a, b, j, 2 * i, ia, mid) + ans(a, b, j, 2 * i + 1, mid + 1, ib)) %
         MOD;
}
long long delta(int a, int b, int j, int i = 1, int ia = 0, int ib = N - 1) {
  if (b < ia || ib < a) return 0;
  if (a <= ia && ib <= b) {
    if (ib <= j) {
      return mults[i];
    } else if (ia > j) {
      return -mults[i];
    }
  }
  int mid = (ia + ib) >> 1;
  return delta(a, b, j, 2 * i, ia, mid) +
         delta(a, b, j, 2 * i + 1, mid + 1, ib);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; ++i) cin >> places[i];
  for (int i = 0; i < n; ++i) cin >> weights[i];
  for (int i = 0; i < n; ++i) setWeight(i, weights[i]);
  for (int t = 0; t < q; ++t) {
    int x, y;
    cin >> x >> y;
    if (x > 0) {
      --x;
      --y;
      int split_low = x;
      int split_high = y;
      while (split_low != split_high) {
        int split_mid = (split_low + split_high) >> 1;
        if (delta(x, y, split_mid) < 0) {
          split_low = split_mid + 1;
        } else {
          split_high = split_mid;
        }
      }
      long long res = ans(x, y, split_low);
      if (res < 0) res += MOD;
      cout << res << '\n';
    } else {
      x = -x;
      --x;
      setWeight(x, y);
    }
  }
}