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;
using lli = long long;
using lld = long double;
using ulli = unsigned long long int;
using pll = pair<lli, lli>;
using ttt = pair<lli, pll>;
using vttt = vector<ttt>;
using vll = vector<pll>;
using vl = vector<lli>;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using cd = complex<lld>;
const double PI = acos(-1);
lli pref[200010], lcol[200010], rcol[200010];
lli arr[200010], wrr[200010];
void add(lli tre[], int x, lli val, bool fl) {
  for (x += 2; x < 200010; x += (x & -x)) {
    tre[x] += val;
    if (fl) tre[x] %= 1000000007;
  }
}
lli get(lli tre[], int x, bool fl) {
  lli res = 0;
  for (x += 2; x > 0; x -= (x & -x)) res += tre[x];
  if (fl) res %= 1000000007;
  return res;
}
lli get(lli tre[], int l, int r, bool fl) {
  if (!fl)
    return get(tre, r, fl) - get(tre, l - 1, fl);
  else
    return (get(tre, r, fl) - get(tre, l - 1, fl) + 1000000007) % 1000000007;
}
lli fin(lli pos, int l, int r) {
  lli move = (get(lcol, pos + 1, r, 1) + get(rcol, l, pos - 1, 1)) % 1000000007;
  lli del1 =
      ((pos - arr[pos]) * (get(pref, l, pos - 1, 0) % 1000000007)) % 1000000007;
  lli del2 =
      ((arr[pos] - pos) * (get(pref, pos + 1, r, 0) % 1000000007)) % 1000000007;
  move = move - del1 - del2 + 1000000007 + 1000000007;
  return move % 1000000007;
}
int tejas_919(int kkkk) {
  lli n, m, k, q, u, v, temp = 0, ans = 0;
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> arr[i];
  for (int i = 0; i < n; i++) {
    cin >> wrr[i];
    add(pref, i, wrr[i], 0);
  }
  {
    for (int i = 0; i < n; i++) {
      temp = (i - arr[i]);
      temp *= wrr[i];
      add(rcol, i, temp, 1);
    }
    for (int i = 0; i < n; i++) {
      temp = (arr[i] - i);
      temp *= wrr[i];
      add(lcol, i, temp, 1);
    }
  }
  for (int i = 1; i < q + 1; i++) {
    cin >> u >> v;
    if (u < 0) {
      u = -u - 1;
      temp = v - wrr[u];
      add(pref, u, temp, 0);
      lli t1 = (temp * (u - arr[u]));
      add(rcol, u, t1, 1);
      t1 = (temp * (arr[u] - u));
      add(lcol, u, t1, 1);
      wrr[u] = v;
    } else {
      if (u == v) {
        cout << 0 << '\n';
        continue;
      } else {
        u--;
        v--;
        lli f = get(pref, u, u, 0);
        lli s = get(pref, u + 1, v, 0);
        if (f >= s) {
          cout << fin(u, u, v) << '\n';
          continue;
        }
        int l = u, r = v;
        f += s;
        f /= 2;
        while (r - l > 1) {
          int mid = (l + r) / 2;
          if (get(pref, u, mid, 0) <= f)
            l = mid;
          else
            r = mid;
        }
        cout << fin(l + 1, u, v) << '\n';
      }
    }
  }
  return 0;
}
signed main() {
  if (!0) {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
  }
  cout << fixed << setprecision(10);
  int t = 1;
  for (int i = 0; i < t; i++) {
    tejas_919(i + 1);
  }
  {};
  if (0) system("pause");
}