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 N = 1000000;
const int MOD = 1000000007;
int le[N], ri[N], a[N];
long long sw[N], swa[N], w[N];
void init(int x, int l, int r) {
  le[x] = l;
  ri[x] = r;
  if (l == r) {
    sw[x] = w[l];
    swa[x] = w[l] * a[l] % MOD;
    return;
  }
  int m = (l + r) >> 1;
  init(x * 2, l, m);
  init(x * 2 + 1, m + 1, r);
  sw[x] = sw[x * 2] + sw[x * 2 + 1];
  swa[x] = (swa[x * 2] + swa[x * 2 + 1]) % MOD;
}
void update(int x, int id, long long nw) {
  if (le[x] == ri[x]) {
    sw[x] = nw;
    swa[x] = nw * a[id] % MOD;
    return;
  }
  int m = (le[x] + ri[x]) >> 1;
  if (id <= m)
    update(x * 2, id, nw);
  else
    update(x * 2 + 1, id, nw);
  sw[x] = sw[x * 2] + sw[x * 2 + 1];
  swa[x] = (swa[x * 2] + swa[x * 2 + 1]) % MOD;
}
long long getsw(int x, int l, int r) {
  if (le[x] >= l && ri[x] <= r) return sw[x];
  int m = (le[x] + ri[x]) >> 1;
  long long tmp = 0;
  if (l <= m) tmp += getsw(x * 2, l, r);
  if (r > m) tmp += getsw(x * 2 + 1, l, r);
  return tmp;
}
int find(int x, int l, int r, long long goalsw, long long &cursw) {
  if (le[x] > r || ri[x] < l) return 0;
  if (le[x] >= l && ri[x] <= r) {
    if (cursw + sw[x] <= goalsw) {
      cursw += sw[x];
      return 0;
    } else if (le[x] == ri[x])
      return le[x];
  }
  int ans = find(x * 2, l, r, goalsw, cursw);
  if (ans) return ans;
  return find(x * 2 + 1, l, r, goalsw, cursw);
}
long long solve(int x, int l, int r, int k) {
  if (le[x] >= l && ri[x] <= r)
    return (sw[x] % MOD * a[k] % MOD - swa[x]) % MOD;
  int m = (le[x] + ri[x]) >> 1;
  long long ret = 0;
  if (l <= m) ret += solve(x * 2, l, r, k);
  if (r > m) ret += solve(x * 2 + 1, l, r, k);
  return ret % MOD;
}
int main() {
  ios::sync_with_stdio(false);
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] -= i;
  }
  for (int i = 1; i <= n; i++) cin >> w[i];
  init(1, 1, n);
  while (q--) {
    int x;
    cin >> x;
    if (x < 0) {
      int id = -x;
      long long nw;
      cin >> nw;
      update(1, id, nw);
    } else {
      int l = x, r;
      cin >> r;
      long long sumw = getsw(1, l, r), cursw = 0;
      int k = find(1, l, r, sumw / 2, cursw);
      long long ans = 0;
      if (l < k) ans += solve(1, l, k - 1, k);
      if (r > k) ans -= solve(1, k + 1, r, k);
      cout << (ans % MOD + MOD) % MOD << "\n";
    }
  }
}