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;
inline bool isvowel(char c) {
  c = tolower(c);
  if (c == 'a' || c == 'e' || c == 'i' || c == 'y' || c == 'o' || c == 'u')
    return 1;
  return 0;
}
const double eps = 0.000001;
const long double pi = acos(-1);
const int maxn = 1e7 + 9;
const int mod = 1e9 + 7;
const long long MOD = 1e18 + 9;
const long long INF = 1e18 + 123;
const int inf = 2e9 + 11;
const int mxn = 1e6 + 9;
const int N = 2e5 + 123;
const int M = mod;
const int pri = 997;
const int Magic = 2101;
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
int n, q;
long long a[N], w[N];
long long f_w[N];
long long f_aw[N];
void inc_w(int x, int y) {
  for (; x < N; x = (x | (x + 1))) {
    f_w[x] += y;
  }
}
void inc_aw(int x, long long y) {
  if (y < 0) y += mod;
  for (; x < N; x = (x | (x + 1))) {
    f_aw[x] += y;
    if (f_aw[x] >= mod) f_aw[x] -= mod;
  }
}
long long get_w(int x) {
  long long res = 0;
  for (; x >= 0; x = (x & (x + 1)) - 1) {
    res += f_w[x];
  }
  return res;
}
int get_aw(int x) {
  int res = 0;
  for (; x >= 0; x = (x & (x + 1)) - 1) {
    res += f_aw[x];
    if (res >= M) res -= M;
  }
  return res;
}
long long get_w(int l, int r) { return get_w(r) - get_w(l - 1); }
int get_aw(int l, int r) {
  int res = get_aw(r);
  res -= get_aw(l - 1);
  if (res < 0) res += M;
  return res;
}
int main() {
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i <= n; i++) {
    a[i] -= i;
    cin >> w[i];
    inc_w(i, w[i]);
    inc_aw(i, a[i] * w[i] % mod);
  }
  for (long long i = 1; i <= q; i++) {
    int x, y;
    cin >> x >> y;
    if (x < 0) {
      x = -x;
      inc_w(x, -w[x]);
      inc_aw(x, -((a[x] * w[x]) % mod));
      w[x] = y;
      inc_w(x, w[x]);
      inc_aw(x, (a[x] * w[x]) % mod);
    } else {
      long long sum = get_w(x, y) / 2;
      int l = x - 1, r = y - 1, p = x;
      while (l <= r) {
        int mid = (l + r) / 2;
        if (get_w(x, mid) <= sum) {
          l = p = mid + 1;
        } else {
          r = mid - 1;
        }
      }
      long long ans = get_aw(p + 1, y) - get_aw(x, p);
      if (ans < 0) ans += mod;
      ans += get_w(x, p) % mod * a[p] % mod;
      ans %= mod;
      ans -= (get_w(p + 1, y) % mod * a[p]) % mod;
      if (ans < 0) ans += mod;
      cout << ans << '\n';
    }
  }
  return 0;
}