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;
template <int sz>
struct FT {
  array<long long, sz + 1> v;
  FT() { fill(v.begin(), v.end(), 0); }
  void update(int i, const long long a) {
    for (++i; i < (int)v.size(); i += i & -i) {
      v[i] += a;
    }
  }
  long long query(int i) const {
    assert(i >= -1 && i < (int)v.size() - 1);
    long long sum = 0;
    for (++i; i; i -= i & -i) {
      sum += v[i];
    }
    return sum;
  }
  long long query(int i, int j) const { return query(j) - query(i - 1); }
};
const int mod = 1e9 + 7;
const int MaxN = (int)2e5;
int n, q;
int a[MaxN];
long long w[MaxN];
FT<MaxN> ftw;
FT<MaxN> ftp;
void setW(const int i, const long long val) {
  assert(0 <= i && i < n);
  w[i] = val;
  ftw.update(i, (-ftw.query(i, i) + val));
  ftp.update(i, (-ftp.query(i, i) + val * (a[i] - i)) % mod);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < n; ++i) {
    long long x;
    cin >> x;
    setW(i, x);
  }
  while (q--) {
    int x, y;
    cin >> x >> y;
    if (x < 0)
      setW(-x - 1, y);
    else {
      --x, --y;
      int lo = x;
      int hi = y;
      const long long total = ftw.query(x, y);
      while (lo < hi) {
        const int mid = (lo + hi) >> 1;
        if (ftw.query(x, mid) >= ((total + 1) >> 1))
          hi = mid;
        else
          lo = mid + 1;
      }
      assert(lo == hi);
      const int k = lo;
      const long long sumL = ftw.query(x, k - 1);
      const long long sumR = ftw.query(k + 1, y);
      long long ans = 0;
      ans += sumL % mod * (a[k] - k) % mod - ftp.query(x, k - 1);
      ans += sumR % mod * (k - a[k]) % mod + ftp.query(k + 1, y);
      ((ans %= mod) += mod) %= mod;
      cout << ans << '\n';
    }
  }
  return 0;
}