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 = 1e9 + 7;
const int MAX = 2e5 + 5;
int pos[MAX], w[MAX], wa_bit[MAX];
long long w_bit[MAX];
int n, q;
inline int add(int a, int b) {
  a += b;
  if (a >= MOD) {
    a -= MOD;
  }
  return a;
}
inline int sub(int a, int b) {
  a -= b;
  if (0 > a) {
    a += MOD;
  }
  return a;
}
inline int mult(int a, int b) { return (1LL * a * b) % MOD; }
void update(int idx, int val, int bit[]) {
  while (idx < MAX) {
    bit[idx] = add(bit[idx], val);
    idx += idx & -idx;
  }
}
int query(int idx, int bit[]) {
  int total = 0;
  while (idx) {
    total = add(total, bit[idx]);
    idx -= idx & -idx;
  }
  return total;
}
int query(int left, int right, int bit[]) {
  if (left > right) {
    return 0;
  }
  return sub(query(right, bit), query(left - 1, bit));
}
void ll_update(int idx, long long val, long long bit[]) {
  while (idx < MAX) {
    bit[idx] += val;
    idx += idx & -idx;
  }
}
long long ll_query(int idx, long long bit[]) {
  long long total = 0;
  while (idx) {
    total += bit[idx];
    idx -= idx & -idx;
  }
  return total;
}
long long ll_query(int left, int right, long long bit[]) {
  if (left > right) {
    return 0;
  }
  return ll_query(right, bit) - ll_query(left - 1, bit);
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = int(1); i < int(n + 1); i++) {
    scanf("%d", pos + i);
    pos[i] -= i;
  }
  for (int i = int(1); i < int(n + 1); i++) {
    scanf("%d", w + i);
    ll_update(i, w[i], w_bit);
    update(i, mult(w[i], pos[i]), wa_bit);
  }
  for (int t = int(0); t < int(q); t++) {
    int x, y;
    scanf("%d %d", &x, &y);
    if (x < 0) {
      int id = -x;
      int nw = y;
      ll_update(id, -w[id], w_bit);
      update(id, (-mult(w[id], pos[id])) + MOD, wa_bit);
      w[id] = nw;
      ll_update(id, w[id], w_bit);
      update(id, mult(w[id], pos[id]), wa_bit);
    } else {
      int l = x;
      int r = y;
      int left = l, right = r;
      int k = -1;
      long long total_sum = ll_query(l, r, w_bit);
      while (left <= right) {
        int mid = (left + right) / 2;
        long long cur_sum = ll_query(l, mid, w_bit);
        if (2 * cur_sum >= total_sum) {
          k = mid;
          right = mid - 1;
        } else {
          left = mid + 1;
        }
      }
      if (k > left) {
        cout << "OI" << endl;
        k--;
      }
      int ans = mult(ll_query(l, k - 1, w_bit) % MOD, pos[k]);
      ans = sub(ans, query(l, k - 1, wa_bit));
      ans = add(ans, query(k + 1, r, wa_bit));
      ans = sub(ans, mult(ll_query(k + 1, r, w_bit) % MOD, pos[k]));
      printf("%d\n", ans);
    }
  }
  return 0;
}