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 maxn = 2e5 + 1;
const int mod = 1e9 + 7;
long long f[2][maxn];
void upd(int r, long long x, int ix = 0) {
  for (; r < maxn; r += r & -r) {
    f[ix][r] += x;
  }
}
long long getf(int r, int ix) {
  long long ans = 0;
  for (; r > 0; r -= r & -r) {
    ans += f[ix][r];
  }
  return ans;
}
long long get(int l, int r, int ix = 0) {
  if (l > r) return 0;
  long long ans = getf(r, ix) - getf(l - 1, ix);
  return ans;
}
long long getm(int l, int r, int ix = 0) { return get(l, r, ix) % mod; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  long long a[n + 1], w[n + 1];
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] -= i;
  }
  for (int i = 1; i <= n; i++) {
    cin >> w[i];
    upd(i, w[i], 0);
    upd(i, a[i] * w[i] % mod, 1);
  }
  while (q--) {
    long long l, r;
    cin >> l >> r;
    if (l < 0) {
      upd(-l, -w[-l] + r);
      upd(-l, (-a[-l] * w[-l]) % mod, 1);
      upd(-l, (a[-l] * r) % mod, 1);
      w[-l] = r;
    } else {
      long long lo = l - 1, hi = r, mid = l;
      long long s = get(l, r);
      while (lo < hi) {
        int mid = (lo + hi) / 2;
        long long s1 = get(l, mid);
        if (s1 * 2 <= s)
          lo = mid + 1;
        else
          hi = mid;
      }
      long long ansl =
          (((a[lo] * getm(l, lo - 1) - getm(l, lo - 1, 1)) % mod) + mod) % mod;
      long long ansr =
          (((getm(lo + 1, r, 1) - a[lo] * getm(lo + 1, r)) % mod) + mod) % mod;
      cout << (ansl + ansr) % mod << '\n';
    }
  }
  return 0;
}