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;
typedef struct _Box {
  int pos, weight;
} Box;
Box arr[200010];
long long int bit1[200010], bit2[200010];
const int mod = 1e9 + 7;
int num;
int getLowbit(int n) { return n & (-n); }
void add(long long int* bitArr, int pos, long long int val, bool shouldMod) {
  if (shouldMod) val = (val + mod) % mod;
  for (int i = pos; i <= num; i += getLowbit(i)) {
    bitArr[i] += val;
    if (shouldMod) bitArr[i] = (bitArr[i] + mod) % mod;
  }
}
long long int getPrefixSum(long long int* bitArr, int pos, bool shouldMod) {
  long long int ans = 0;
  for (int i = pos; i > 0; i -= getLowbit(i)) {
    ans += bitArr[i];
    if (shouldMod) ans = (ans + mod) % mod;
  }
  return ans;
}
long long int getRangeSum(long long int* bitArr, int qLeftPt, int qRightPt,
                          bool shouldMod) {
  if (qRightPt < qLeftPt) return 0;
  long long int ans = getPrefixSum(bitArr, qRightPt, shouldMod) -
                      getPrefixSum(bitArr, qLeftPt - 1, shouldMod);
  if (shouldMod) ans = (ans + mod) % mod;
  return ans;
}
int getMidPos(int qLeftPt, int qRightPt) {
  int leftPt = qLeftPt, rightPt = qRightPt;
  int ans = rightPt;
  while (leftPt <= rightPt) {
    int midPt = (leftPt + rightPt) >> 1;
    if (getRangeSum(bit1, qLeftPt, midPt, false) >=
        getRangeSum(bit1, midPt + 1, qRightPt, false)) {
      ans = midPt;
      rightPt = midPt - 1;
    } else {
      leftPt = midPt + 1;
    }
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  memset(bit1, 0, sizeof(bit1));
  memset(bit2, 0, sizeof(bit2));
  int qNum;
  cin >> num >> qNum;
  for (int i = 1; i <= num; i++) {
    cin >> arr[i].pos;
  }
  for (int i = 1; i <= num; i++) {
    cin >> arr[i].weight;
    add(bit1, i, arr[i].weight, false);
    add(bit2, i, (long long int)arr[i].weight * (arr[i].pos - i), true);
  }
  while (qNum--) {
    int x, y;
    cin >> x >> y;
    if (x < 0) {
      x = -x;
      add(bit1, x, y - arr[x].weight, false);
      add(bit2, x, ((long long int)y - arr[x].weight) * (arr[x].pos - x), true);
      arr[x].weight = y;
    } else {
      if (x == y) {
        cout << 0 << endl;
        continue;
      }
      int midPos = getMidPos(x, y);
      long long int leftAns = (getRangeSum(bit1, x, midPos, true) *
                                   abs(arr[midPos].pos - midPos) % mod -
                               getRangeSum(bit2, x, midPos, true) + mod) %
                              mod;
      long long int rightAns = (getRangeSum(bit1, midPos, y, true) *
                                    abs(arr[midPos].pos - midPos) % mod -
                                getRangeSum(bit2, midPos, y, true) + mod) %
                               mod;
      long long int ans = (leftAns - rightAns + mod) % mod;
      cout << ans << endl;
    }
  }
  return 0;
}