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;
inline void addm(long long& a, const long long& b, const long long& mod) {
  a += b;
  while (a < 0) a += mod;
  while (a >= mod) a -= mod;
}
const long long mod = 1e9 + 7;
class fenwick_tree {
  long long mymod;
  int n;
  vector<long long> v;

 public:
  fenwick_tree(int _n, long long _mod)
      : n(_n), mymod(_mod), v(vector<long long>(n + 2, 0)) {}
  void add(int in, long long val) {
    for (; in < n; in = in | (in + 1)) addm(v[in], val, mymod);
  }
  long long sum(int r) {
    long long rval = 0;
    for (; r >= 0; r = (r & (r + 1)) - 1) addm(rval, v[r], mymod);
    return rval;
  }
  long long sum(int l, int r) {
    if (r < l) return 0;
    long long res = sum(r);
    addm(res, -(!l ? 0 : sum(l - 1)), mymod);
    return res;
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  vector<long long> a(n), w(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  fenwick_tree sum(n, 1e18), wa(n, mod), wi(n, mod);
  for (int i = 0; i < n; i++) {
    cin >> w[i];
    sum.add(i, w[i]);
    wa.add(i, (a[i] * w[i]) % mod);
    wi.add(i, (i * 1ll * w[i]) % mod);
  }
  while (q--) {
    long long x, y;
    cin >> x >> y;
    if (x < 0) {
      x = -x - 1;
      sum.add(x, y - w[x]);
      wa.add(x, (((mod + y - w[x]) % mod) * a[x]) % mod);
      wi.add(x, (((mod + y - w[x]) % mod) * 1ll * x) % mod);
      w[x] = y;
    } else {
      long long s = sum.sum(--x, --y);
      int lo = x, hi = y;
      while (lo != hi) {
        int mid = (lo + hi) / 2 + 1;
        if (sum.sum(x, mid - 1) * 2 <= s)
          lo = mid;
        else
          hi = mid - 1;
      }
      long long sumwleft = sum.sum(x, lo) % mod;
      long long numleft = lo - x + 1;
      long long sumawleft = wa.sum(x, lo) % mod;
      long long sumwileft = wi.sum(x, lo) % mod;
      long long totalsumleft = (a[lo] * sumwleft) % mod - sumawleft -
                               (((sumwleft * lo) % mod) - sumwileft) % mod;
      addm(totalsumleft, 0, mod);
      long long sumwright = sum.sum(lo, y) % mod;
      long long numright = y - lo + 1;
      long long sumawright = wa.sum(lo, y) % mod;
      long long sumwiright = wi.sum(lo, y) % mod;
      long long totalsumright = sumawright - (a[lo] * sumwright) % mod -
                                (sumwiright - ((sumwright * lo) % mod)) % mod;
      addm(totalsumright, 0, mod);
      cout << (totalsumleft + totalsumright) % mod << "\n";
    }
  }
  return 0;
}