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 long long MOD = 1e9 + 7;
const long long BIG = 1e9 + 5;
int n, q;
long long A[200003], W[200003];
long long ft[200003];
long long sum(int a) { return (a) ? ft[a] + sum(a - (a & -a)) : 0; }
long long sum(int a, int b) { return (a > b) ? 0 : sum(b) - sum(a - 1); }
void update(int a, long long val) {
  for (; a <= n; a += (a & -a)) ft[a] += val;
}
long long ftk[200003][2];
long long sumk(int a, int tipe) {
  return (a) ? (ftk[a][tipe] + sumk(a - (a & -a), tipe)) % MOD : 0;
}
long long sumk(int a, int b, int tipe) {
  return (a > b) ? 0 : sumk(b, tipe) - sumk(a - 1, tipe);
}
void updatek(int a, long long val, int tipe) {
  for (; a <= n; a += (a & -a)) ftk[a][tipe] = (ftk[a][tipe] + val) % MOD;
}
int binser(int l, int r, int x) {
  int iki = l, ika = r;
  while (iki < ika) {
    int imi = (1 + iki + ika) / 2;
    if (A[imi] > x)
      ika = imi - 1;
    else
      iki = imi;
  }
  return iki;
}
bool cek(int l, int r, int x) {
  int idx = binser(l, r, x);
  return sum(l, idx) <= sum(idx + 1, r);
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%lld", &A[i]);
  for (int i = 1; i <= n; i++) scanf("%lld", &W[i]);
  for (int i = 1; i <= n; i++) {
    update(i, W[i]);
    updatek(i, ((A[i] - i) * W[i]) % MOD, 0);
    updatek(i, ((-A[i] + i + BIG) * W[i]) % MOD, 1);
  }
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    if (l < 0) {
      l = -l;
      update(l, -W[l]);
      updatek(l, -((A[l] - l) * W[l]) % MOD, 0);
      updatek(l, -((-A[l] + l + BIG) * W[l]) % MOD, 1);
      W[l] = r;
      update(l, W[l]);
      updatek(l, ((A[l] - l) * W[l]) % MOD, 0);
      updatek(l, ((-A[l] + l + BIG) * W[l]) % MOD, 1);
    } else {
      long long kir = A[l], kan = A[r];
      while (kir < kan) {
        long long mid = (1 + kir + kan) / 2;
        if (!cek(l, r, mid))
          kan = mid - 1;
        else
          kir = mid;
      }
      int idx = binser(l, r, kir);
      long long res = 0;
      res +=
          (sumk(l, idx, 1) - ((BIG + idx - kir) * (sum(l, idx) % MOD)) % MOD) %
          MOD;
      res += (sumk(idx + 1, r, 0) -
              ((kir - idx) * (sum(idx + 1, r) % MOD)) % MOD) %
             MOD;
      res %= MOD;
      printf("%lld\n", (res % MOD + MOD) % MOD);
    }
  }
}