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 = 1000000007;
class fenwicktree {
 private:
  int sz, realsz;
  vector<long long> val;

 public:
  fenwicktree() : sz(0), val(vector<long long>(1, 0)){};
  fenwicktree(int sz_) : realsz(sz_) {
    sz = 1;
    while (sz <= sz_) sz *= 2;
    val = vector<long long>(sz + 1, 0);
  }
  void add(int pos, long long delta) {
    for (int i = pos + 1; i <= sz; i += i & (-i)) {
      val[i] += delta;
    }
  }
  long long sum(int pos) {
    long long ans = 0;
    for (int i = pos; i >= 1; i -= i & (-i)) {
      ans += val[i];
    }
    return ans;
  }
  int bsearch(long long x) {
    if (x == 0) return 0;
    int ans = 0;
    for (int i = sz / 2; i >= 1; i >>= 1) {
      if (val[ans + i] < x) {
        ans += i;
        x -= val[ans];
      }
    }
    if (ans > realsz) ans = realsz;
    return ans + 1;
  }
};
class fenwicktree_mod {
 private:
  int sz, realsz;
  vector<int> val;

 public:
  fenwicktree_mod() : sz(0), val(vector<int>(1, 0)){};
  fenwicktree_mod(int sz_) : sz(sz_) { val = vector<int>(sz + 1, 0); }
  void add(int pos, long long delta) {
    if (delta < 0) delta = (mod - (-delta) % mod);
    delta %= mod;
    for (int i = pos + 1; i <= sz; i += i & (-i)) {
      val[i] += delta;
      if (val[i] >= mod) val[i] -= mod;
    }
  }
  int sum(int pos) {
    int ans = 0;
    for (int i = pos; i >= 1; i -= i & (-i)) {
      ans += val[i];
      if (ans >= mod) ans -= mod;
    }
    return ans;
  }
};
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  int N, Q;
  cin >> N >> Q;
  vector<int> A(N), W(N);
  for (int i = 0; i < N; ++i) cin >> A[i], A[i] -= i;
  for (int i = 0; i < N; ++i) cin >> W[i];
  fenwicktree seg(N);
  fenwicktree_mod sum(N);
  for (int i = 0; i < N; ++i) {
    seg.add(i, W[i]);
    sum.add(i, 1LL * W[i] * A[i]);
  }
  for (int i = 0; i < Q; ++i) {
    int x, y;
    cin >> x >> y;
    if (x < 0) {
      x *= -1;
      --x;
      seg.add(x, -W[x]);
      sum.add(x, -1LL * W[x] * A[x]);
      W[x] = y;
      seg.add(x, W[x]);
      sum.add(x, 1LL * W[x] * A[x]);
    } else {
      --x;
      long long sl = seg.sum(x);
      long long sr = seg.sum(y);
      int med = seg.bsearch((sl + sr + 1) / 2);
      int suml = sum.sum(x);
      int summ = sum.sum(med);
      int sumr = sum.sum(y);
      long long sm = seg.sum(med);
      int lcost =
          (1LL * (sm - sl) % mod * A[med - 1] - (summ - suml) + mod) % mod;
      int rcost = ((sumr - summ) - 1LL * (sr - sm) % mod * A[med - 1] % mod +
                   2LL * mod) %
                  mod;
      cout << (lcost + rcost) % mod << '\n';
    }
  }
  return 0;
}