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 nmax = 200005;
const long long mod = 1000 * 1000 * 1000 + 7;
long long aib[nmax], fft[nmax], a[nmax], w[nmax];
int n, i, j, q, tip;
long long l, r, poz, val, opt, ans, aa, b, tot;
inline int lbit(int x) { return (((x ^ (x - 1)) & x)); }
void update(long long poz, long long val) {
  long long vv = ((1LL * a[poz] * val) % mod + mod) % mod;
  for (int idx = poz; idx <= n; idx += lbit(idx)) {
    aib[idx] = aib[idx] + val;
    fft[idx] = (fft[idx] + vv);
    fft[idx] = fft[idx] - mod * (fft[idx] >= mod);
  }
}
long long compute(int poz, int p) {
  long long ret1 = 0, ret2 = 0;
  for (int idx = poz; idx > 0; idx -= lbit(idx)) {
    ret1 += aib[idx];
    ret2 = (ret2 + fft[idx]);
    ret2 = ret2 - mod * (ret2 >= mod);
  }
  if (!p) return ret1;
  return ret2;
}
int find_kth(long long x) {
  int ret = 0;
  long long avem = 0;
  for (int p = 17; p >= 0; p--)
    if (ret + (1 << p) < n && avem + aib[ret + (1 << p)] <= x) {
      ret += (1 << p);
      avem += aib[ret];
    }
  ret++;
  return ret;
}
bool chk(int val) { return (compute(val - 1, 0) - compute(l - 1, 0) <= tot); }
long long cc(long long x) { return ((1LL * x * (x + 1)) / 2) % mod; }
long long calc(long long opt) {
  ans =
      compute(l - 1, 1) - compute(opt - 1, 1) - compute(opt, 1) + compute(r, 1);
  ans = (ans + 2 * mod) % mod;
  ans = (1LL * ans +
         1LL * ((compute(opt - 1, 0) - compute(l - 1, 0)) % mod) * a[opt] -
         1LL * ((compute(r, 0) - compute(opt, 0)) % mod) * a[opt]) %
        mod;
  ans = (ans + mod) % mod;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> q;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] -= i;
  }
  for (i = 1; i <= n; i++) {
    cin >> w[i];
    update(i, w[i]);
  }
  long long ret = 0;
  for (i = 1; i <= q; i++) {
    cin >> aa >> b;
    if (aa < 0)
      tip = 1, aa *= -1;
    else
      tip = 2;
    if (tip == 1) {
      poz = aa;
      val = b;
      update(poz, val - w[poz]);
      w[poz] = val;
    } else {
      l = aa;
      r = b;
      ans = 0;
      opt = l;
      opt = find_kth((-compute(l - 1, 0) + compute(r, 0)) / 2);
      tot = (-compute(l - 1, 0) + compute(r, 0)) / 2;
      for (int p = 17; p >= 0; p--)
        if (opt + (1 << p) <= r && chk(opt + (1 << p))) opt += (1 << p);
      ret = calc(opt);
      cout << ret << '\n';
    }
  }
  return 0;
}