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 long long MOD = 1e9 + 7;
const int MAXN = 2e5 + 10;
long long a[MAXN], w[MAXN], conv[MAXN];
struct BIT {
  long long b[MAXN];
  bool mod;
  void upd(int p, long long v) {
    for (; p < MAXN; p += (p & -p)) {
      b[p] += v;
      if (mod) b[p] %= MOD;
    }
  }
  long long sum(int p) {
    long long res = 0;
    for (; p; p -= (p & -p)) {
      res += b[p];
      if (res >= MOD && mod) res -= MOD;
    }
    return res;
  }
  long long sum(int l, int r) {
    if (l > r) return 0;
    return sum(r) - sum(l - 1);
  }
};
BIT wBIT, convBIT, progBIT;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  convBIT.mod = true;
  progBIT.mod = true;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) {
    cin >> w[i];
    wBIT.upd(i, w[i]);
    progBIT.upd(i, (long long)i * w[i]);
    convBIT.upd(i, w[i] * a[i]);
  }
  while (q--) {
    int x, y;
    cin >> x >> y;
    if (x < 0) {
      x *= -1;
      long long d = (long long)y - w[x];
      wBIT.upd(x, d);
      progBIT.upd(x, d * x);
      convBIT.upd(x, d * a[x]);
      w[x] = y;
    } else {
      if (x == y) {
        cout << "0\n";
        continue;
      }
      int lo = x, hi = y;
      while (lo < hi) {
        int mi = (lo + hi + 1) / 2;
        if (wBIT.sum(mi + 1, y) >= wBIT.sum(x, mi))
          lo = mi;
        else
          hi = mi - 1;
      }
      lo++;
      if (wBIT.sum(x + 1, y) < w[x]) lo = x;
      long long ans = 0;
      ans += convBIT.sum(lo + 1, y);
      ans -= progBIT.sum(lo + 1, y);
      ans += (long long)(lo - a[lo]) * (wBIT.sum(lo + 1, y) % MOD);
      ans %= MOD;
      ans -= convBIT.sum(x, lo - 1);
      ans += progBIT.sum(x, lo - 1);
      ans += (long long)(a[lo] - lo) * (wBIT.sum(x, lo - 1) % MOD);
      ans %= MOD;
      cout << (ans + MOD) % MOD << '\n';
    }
  }
}