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 maxn = 200005;
const int mod = 1000000007;
long long a[maxn], w[maxn];
long long pos[maxn];
int n, q;
struct Bit {
  long long sum[maxn];
  int n;
  bool mod_flag;
  Bit(int n_) {
    n = n_;
    mod_flag = true;
    memset(sum, 0, sizeof sum);
  }
  void init(int n_, bool flag = true) {
    n = n_;
    mod_flag = flag;
  }
  void add(int x, long long val) {
    while (x <= n) {
      sum[x] += val;
      if (mod_flag) {
        if (sum[x] < 0) sum[x] += mod;
        if (sum[x] >= mod) sum[x] %= mod;
      }
      x += x & -x;
    }
  }
  long long get(int x) {
    long long res = 0;
    while (x) {
      res += sum[x];
      if (mod_flag) {
        if (res >= mod) res %= mod;
      }
      x -= x & -x;
    }
    return res;
  }
  long long query(int l, int r) {
    long long res = get(r) - get(l - 1);
    if (mod_flag) {
      res = (res + mod) % mod;
    }
    return res;
  }
};
Bit bit1(maxn), bit2(maxn), bit3(maxn);
int main() {
  cin >> n >> q;
  int m = 1e9;
  bit1.init(n, false);
  bit2.init(n);
  bit3.init(n);
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) {
    cin >> w[i];
    pos[i] = m - (n - i);
    bit1.add(i, w[i]);
    bit2.add(i, w[i] * (a[i] - i));
    bit3.add(i, w[i] * (pos[i] - a[i]));
  }
  while (q--) {
    int x, y;
    cin >> x >> y;
    if (x < 0) {
      int id = -x;
      bit1.add(id, -w[id]);
      bit2.add(id, (a[id] - id) * (mod - w[id]));
      bit3.add(id, (pos[id] - a[id]) * (mod - w[id]));
      w[id] = y;
      bit1.add(id, w[id]);
      bit2.add(id, (a[id] - id) * w[id]);
      bit3.add(id, (pos[id] - a[id]) * w[id]);
    } else {
      long long res = bit1.get(y) - bit1.get(x - 1);
      long long middlesum = res / 2;
      int l = x, r = y, opt = -1;
      while (l <= r) {
        int m = (l + r) >> 1;
        if (bit1.get(m) - bit1.get(x - 1) > middlesum)
          r = m - 1, opt = m;
        else
          l = m + 1;
      }
      long long ans = 0;
      if (opt < y) {
        long long diff = a[opt] - opt;
        long long rsum = bit1.query(opt + 1, y) % mod;
        ans = (ans + bit2.query(opt + 1, y) - rsum * diff % mod + mod) % mod;
      }
      if (opt > x) {
        long long diff = pos[opt] - a[opt];
        long long lsum = bit1.query(x, opt - 1) % mod;
        ans = (ans + bit3.query(x, opt - 1) - lsum * diff % mod + mod) % mod;
      }
      cout << ans << endl;
    }
  }
  return 0;
}