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 INF = 1e9 + 7, N = 2e5 + 7, M = 1e9 + 7;
int a[N], w[N];
struct {
  long long bit[N];
  void add(int i, long long val = 1) {
    for (; i < N; i += i & -i) bit[i] += val;
  }
  long long qry(int i) {
    long long res = 0;
    for (; i; i -= i & -i) res += bit[i];
    return res;
  }
} b1;
struct {
  int bit[N];
  void add(int i, int val = 1) {
    for (; i < N; i += i & -i) bit[i] = (bit[i] + val) % M;
  }
  int qry(int i) {
    int res = 0;
    for (; i; i -= i & -i) res = (res + bit[i]) % M;
    return res;
  }
} b2;
signed main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  ;
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] -= i;
  }
  for (int i = 1; i <= n; i++) {
    cin >> w[i];
    b1.add(i, w[i]);
    b2.add(i, (long long)a[i] * w[i] % M);
  }
  while (q--) {
    int x, y;
    cin >> x >> y;
    if (x < 0) {
      int i = -x, nw = y;
      b1.add(i, -w[i]);
      b2.add(i, (M - (long long)a[i] * w[i] % M) % M);
      w[i] = nw;
      b1.add(i, w[i]);
      b2.add(i, (long long)a[i] * w[i] % M);
    } else {
      int l = x, r = y;
      if (l == r) {
        cout << 0 << endl;
        continue;
      }
      int cur = 0;
      long long tot = b1.qry(r) - b1.qry(l - 1), sum = 0, tt = b1.qry(l - 1);
      for (int j = 19; j >= 0; j--) {
        if (cur + (1 << j) <= r) {
          int tmp = cur + (1 << j);
          long long summ = b1.bit[tmp] + sum;
          if ((summ - tt) * 2 < tot) {
            sum = summ;
            cur = tmp;
          }
        }
      }
      int to = cur + 1;
      assert(to <= r);
      long long ans = (b1.qry(to - 1) - b1.qry(l - 1)) % M * a[to] % M;
      ans = (ans + -(b1.qry(r) - b1.qry(to)) % M * a[to] % M % M + M) % M;
      ans = (ans - b2.qry(to - 1) + b2.qry(l - 1) + M) % M;
      ans = (ans + b2.qry(r) - b2.qry(to) + M) % M;
      cout << ans << endl;
    }
  }
}