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;
int N, Q;
int A[200005], W[200005];
long long Arb[2][200005];
const int MOD = 1000000007;
void Read() {
  scanf("%d%d", &N, &Q);
  for (int i = 1; i <= N; i++) {
    scanf("%d", &A[i]);
  }
  for (int i = 1; i <= N; i++) scanf("%d", &W[i]);
}
int norm(int x) {
  if (x < 0) x += MOD;
  return x >= MOD ? x - MOD : x;
}
void update(int ind, int pos, int val) {
  while (pos <= N) {
    if (ind == 0)
      Arb[ind][pos] += val;
    else
      Arb[ind][pos] = norm(Arb[ind][pos] + val);
    pos += (pos & (-pos));
  }
}
long long sum(int ind, int pos) {
  long long ret = 0;
  while (pos >= 1) {
    if (ind == 1)
      ret = norm(ret + Arb[ind][pos]);
    else
      ret += Arb[ind][pos];
    pos -= (pos & (-pos));
  }
  return ret;
}
long long sumBetween(int ind, int from, int to) {
  if (from > to) return 0;
  if (ind == 1)
    return norm(sum(ind, to) + MOD - sum(ind, from - 1));
  else
    return sum(ind, to) - sum(ind, from - 1);
}
int findPoint(int x, int y) {
  int left = x, right = y, mid, sol = left;
  while (left <= right) {
    mid = (left + right) / 2;
    if (sumBetween(0, x, mid) - sumBetween(0, mid + 1, y) >= 0) {
      sol = mid;
      right = mid - 1;
    } else
      left = mid + 1;
  }
  return sol;
}
int getWeight(int pos, int x, int y, int point) {
  int sum1 = sumBetween(1, x, point);
  int ans = 0;
  int sum0 = (1LL * (sumBetween(0, x, point) % MOD) * (x - pos)) % MOD;
  if (sum0 < 0) sum0 += MOD;
  ans = norm(ans + sum1);
  ans = norm(ans + sum0);
  sum1 = sumBetween(1, point + 1, y);
  sum0 = (1LL * (sumBetween(0, point + 1, y) % MOD) * (x - pos)) % MOD;
  if (sum0 < 0) sum0 += MOD;
  ans = norm(ans + MOD - sum1);
  ans = norm(ans + MOD - sum0);
  return norm(MOD - ans);
}
int query(int x, int y) {
  int point = findPoint(x, y);
  int pos = A[point] - (point - x);
  return getWeight(pos, x, y, point);
}
void init() {
  for (int i = 1; i <= N; i++) {
    update(0, i, W[i]);
    update(1, i, norm((1LL * W[i] * (A[i] - i)) % MOD));
  }
}
void pointUpdate(int point, int val) {
  update(0, point, -W[point]);
  int aux = norm((1LL * W[point] * (A[point] - point)) % MOD);
  update(1, point, MOD - aux);
  update(0, point, val);
  update(1, point, norm(1LL * val * (A[point] - point) % MOD));
  W[point] = val;
}
void Solve() {
  for (int i = 1; i <= Q; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x > 0) {
      printf("%d\n", query(x, y));
    } else
      pointUpdate(-x, y);
  }
}
int main() {
  Read();
  init();
  Solve();
  return 0;
}