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;
long long int n, m, a[1000005];
long long int w[1000005];
long long int bit1[1000005];
long long int bit2[1000005];
void upd(long long int idx, long long int val, long long int bit[]) {
  for (; idx < 1000005; idx += idx & -idx) bit[idx] += val;
}
long long int qry(long long int idx, long long int bit[]) {
  long long int ans = 0;
  for (; idx > 0; idx -= idx & -idx) ans += bit[idx];
  return ans;
}
long long int inline mul(long long int a, long long int b) {
  a %= 1000000007;
  b %= 1000000007;
  if (a < 0) a += 1000000007;
  if (b < 0) b += 1000000007;
  a *= b;
  a %= 1000000007;
  return a;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long int q;
  cin >> n >> q;
  for (long long int i = 1; i <= n; i++) cin >> a[i], a[i] -= i;
  for (long long int i = 1; i <= n; i++) cin >> w[i];
  for (long long int i = 1; i <= n; i++) {
    upd(i, mul(w[i], a[i]), bit1);
    upd(i, w[i], bit2);
  }
  for (long long int i = 1; i <= q; i++) {
    long long int l, r;
    cin >> l >> r;
    if (l < 0) {
      l = -l;
      upd(l, -1LL * mul(w[l], a[l]), bit1);
      upd(l, -1LL * w[l], bit2);
      w[l] = r;
      upd(l, mul(w[l], a[l]), bit1);
      upd(l, w[l], bit2);
    } else {
      long long int lo = l;
      long long int hi = r;
      long long int mid = l;
      long long int sm = qry(r, bit2) - qry(l - 1, bit2);
      while (lo <= hi) {
        long long int m = (lo + hi) / 2;
        if (2 * (qry(m - 1, bit2) - qry(l - 1, bit2)) <= sm)
          lo = m + 1, mid = m;
        else
          hi = m - 1;
      }
      long long int ans = mul((qry(mid, bit2) - qry(l - 1, bit2)), a[mid]) -
                          (qry(mid, bit1) - qry(l - 1, bit1)) % 1000000007;
      ans %= 1000000007;
      if (ans < 0) ans += 1000000007;
      long long int ans2 = mul((qry(r, bit2) - qry(mid, bit2)), a[mid]) -
                           (qry(r, bit1) - qry(mid, bit1)) % 1000000007;
      ans2 %= 1000000007;
      ans2 += 1000000007;
      ans2 %= 1000000007;
      ans -= ans2;
      ans %= 1000000007;
      ans += 1000000007;
      ans %= 1000000007;
      cout << ans << '\n';
    }
  }
  return 0;
}