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 double PI = acos(-1.0);
const int INF = 1000 * 1000 * 1000 + 7;
const long long LINF = INF * (long long)INF;
const int MAX = 2 * 1000 * 100 + 47;
const int MOD = 1000 * 1000 * 1000 + 7;
struct Fenvik {
  long long T[MAX];
  int n;
  void init(int n1) { n = n1; }
  void add(int i, int delta) {
    for (; i < n; i = i | (i + 1)) {
      T[i] += delta;
    }
  }
  long long get(int i) {
    long long res = 0;
    for (; i >= 0; i = (i & (i + 1)) - 1) {
      res = res + T[i];
    }
    return res;
  }
  long long get(int l, int r) {
    long long res = get(r);
    if (l) {
      res -= get(l - 1);
    }
    return res;
  }
  int getNorm(int l, int r) {
    long long res = get(l, r) % MOD;
    if (res < 0) res += MOD;
    if (res >= MOD) res -= MOD;
    return res;
  }
} F1, F2;
long long A[MAX];
long long W[MAX];
int n;
void upd(int pos, int val) {
  long long delta = val - W[pos];
  F1.add(pos, val - W[pos]);
  F2.add(pos, delta * (pos - A[pos]) % MOD);
  W[pos] = val;
}
int get(int x, int y) {
  int l = x - 1, r = y;
  long long s = F1.get(x, y);
  while (r - l > 1) {
    int c = (r + l) / 2;
    if (2 * F1.get(x, c) >= s) {
      r = c;
    } else {
      l = c;
    }
  }
  int k = r;
  l = x;
  r = y;
  long long dod1 = (A[k] - k) * F1.getNorm(l, k - 1) + F2.get(l, k - 1);
  long long dod2 = (k - A[k]) * F1.getNorm(k + 1, r) - F2.get(k + 1, r);
  long long res = (dod1 + dod2) % MOD;
  if (res < 0) res += MOD;
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int q;
  cin >> n >> q;
  for (int i = (0); i < (n); i++) cin >> A[i];
  for (int i = (0); i < (n); i++) cin >> W[i];
  F1.init(n);
  F2.init(n);
  for (int i = (0); i < (n); i++) {
    F1.add(i, W[i]);
    F2.add(i, (i - A[i]) * W[i] % MOD);
  }
  for (int i = (0); i < (q); i++) {
    int x, y;
    cin >> x >> y;
    if (x < 0) {
      upd(-x - 1, y);
    } else {
      x--;
      y--;
      int res = get(x, y);
      cout << res << "\n";
    }
  }
}