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;
struct Fenwick {
  vector<long long int> bit;
  long long int size;
  void modify(long long int j, long long int v) {
    j++;
    for (; j < size; j += j & -j) bit[j] += v;
  }
  long long int query(long long int j) {
    j++;
    long long int v = 0;
    for (; j > 0; j -= j & -j) v += bit[j];
    return v;
  }
  long long int rangequery(long long int a, long long int b) {
    if (b < a) return 0;
    return query(b) - query(a - 1);
  }
  void rangeupdate(long long int a, long long int b, long long int v) {
    modify(a, v);
    modify(b + 1, -v);
  }
  Fenwick(long long int s) {
    s += 3;
    size = s;
    bit = vector<long long int>(s, 0);
  }
};
struct FenwickM {
  vector<long long int> bit;
  long long int size;
  void modify(long long int j, long long int v) {
    v %= 1000000007;
    v = (v + 1000000007) % 1000000007;
    j++;
    for (; j < size; j += j & -j) bit[j] = (bit[j] + v) % 1000000007;
  }
  long long int query(long long int j) {
    j++;
    long long int v = 0;
    for (; j > 0; j -= j & -j) v = (bit[j] + v) % 1000000007;
    return v;
  }
  long long int rangequery(long long int a, long long int b) {
    return (query(b) - query(a - 1) + 1000000007) % 1000000007;
  }
  void rangeupdate(long long int a, long long int b, long long int v) {
    modify(a, v);
    modify(b + 1, -v);
  }
  FenwickM(long long int s) {
    s += 3;
    size = s;
    bit = vector<long long int>(s, 0);
  }
};
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, q;
  cin >> n >> q;
  vector<long long int> X(n + 1);
  vector<long long int> W(n + 1);
  vector<pair<long long int, long long int> > Q(q);
  vector<long long int> beg;
  Fenwick pref(200005);
  FenwickM data(200005);
  for (long long int i = 1; i <= n; i++) cin >> X[i];
  for (long long int i = 1; i <= n; i++) cin >> W[i];
  beg = W;
  for (long long int i = 0; i < q; i++) {
    cin >> Q[i].first >> Q[i].second;
  }
  for (long long int i = 1; i <= n; i++) {
    pref.modify(i, W[i]);
    data.modify(i, (i - X[i]) * W[i]);
  }
  for (long long int k = 0; k < q; k++) {
    if (Q[k].first < 0) {
      long long int i = -Q[k].first;
      pref.modify(i, Q[k].second - W[i]);
      data.modify(i, -(i - X[i]) * W[i]);
      data.modify(i, (i - X[i]) * Q[k].second);
      W[i] = Q[k].second;
    } else {
      long long int L = Q[k].first, R = Q[k].second;
      long long int l = L, r = R;
      long long int med = L - 1;
      if (L == R) {
        cout << "0\n";
        continue;
      }
      while (l <= r) {
        long long int mid = (l + r) / 2;
        if (pref.rangequery(L, mid) <= pref.rangequery(mid + 1, R)) {
          med = mid;
          l = mid + 1;
        } else
          r = mid - 1;
      }
      long long int ans = 0;
      if (med >= L)
        ans = (pref.rangequery(L, med) % 1000000007 * (X[med + 1] - med - 1) %
                   1000000007 +
               data.rangequery(L, med)) %
              1000000007;
      med += 2;
      if (med <= R)
        ans = (ans -
               pref.rangequery(med, R) % 1000000007 * (X[med - 1] - med + 1) %
                   1000000007 -
               data.rangequery(med, R)) %
              1000000007;
      cout << (ans % 1000000007 + 1000000007) % 1000000007 << "\n";
    }
  }
}