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 MX = 200005;
const int MD = 1000000007;
int a[MX];
int w[MX];
struct {
  int L, R;
  int pl, pr;
  long long sum, sum2;
} T[2 * MX];
int buildTree(int L, int R) {
  static int tsz = 0;
  int cur = tsz++;
  T[cur].L = L;
  T[cur].R = R;
  if (L < R) {
    T[cur].pl = buildTree(L, (L + R) / 2);
    T[cur].pr = buildTree((L + R) / 2 + 1, R);
    T[cur].sum = T[T[cur].pl].sum + T[T[cur].pr].sum;
    T[cur].sum2 = T[T[cur].pl].sum2 + T[T[cur].pr].sum2;
    T[cur].sum2 %= MD;
  } else {
    T[cur].sum = w[L];
    T[cur].sum2 = (w[L] * 1LL * a[L]) % MD;
  }
  return cur;
}
long long st_S;
int get(int cur, int L) {
  if (L > T[cur].R) return -1;
  if (T[cur].L == T[cur].R) {
    st_S -= T[cur].sum;
    if (st_S <= 0) {
      return T[cur].L;
    }
    return -1;
  }
  if (L >= T[T[cur].pr].L) return get(T[cur].pr, L);
  if (L >= T[cur].L) {
    int tmp = get(T[cur].pl, L);
    if (tmp != -1) return tmp;
    return get(T[cur].pr, L);
  }
  if (T[cur].sum < st_S) {
    st_S -= T[cur].sum;
    return -1;
  }
  int tmp = get(T[cur].pl, L);
  if (tmp != -1) return tmp;
  return get(T[cur].pr, L);
}
long long getSum(int cur, int L, int R) {
  if (L > R) return 0;
  if (L <= T[cur].L && R >= T[cur].R) return T[cur].sum;
  long long sum = 0;
  if (L <= T[T[cur].pl].R && T[T[cur].pl].L <= R) {
    sum += getSum(T[cur].pl, L, R);
  }
  if (L <= T[T[cur].pr].R && T[T[cur].pr].L <= R) {
    sum += getSum(T[cur].pr, L, R);
  }
  return sum;
}
long long getSum2(int cur, int L, int R) {
  if (L > R) return 0;
  if (L <= T[cur].L && R >= T[cur].R) return T[cur].sum2;
  long long sum2 = 0;
  if (L <= T[T[cur].pl].R && T[T[cur].pl].L <= R) {
    sum2 += getSum2(T[cur].pl, L, R);
  }
  if (L <= T[T[cur].pr].R && T[T[cur].pr].L <= R) {
    sum2 += getSum2(T[cur].pr, L, R);
  }
  sum2 %= MD;
  return sum2;
}
void upd(int cur, int i) {
  if (T[cur].L == T[cur].R) {
    T[cur].sum = w[i];
    T[cur].sum2 = (w[i] * 1LL * a[i]) % MD;
    return;
  }
  if (i <= T[T[cur].pl].R)
    upd(T[cur].pl, i);
  else
    upd(T[cur].pr, i);
  T[cur].sum = T[T[cur].pl].sum + T[T[cur].pr].sum;
  T[cur].sum2 = T[T[cur].pl].sum2 + T[T[cur].pr].sum2;
  T[cur].sum2 %= MD;
}
int main() {
  int n, q;
  ignore = scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) {
    ignore = scanf("%d", a + i);
    a[i] -= i;
  }
  for (int i = 1; i <= n; i++) {
    ignore = scanf("%d", w + i);
  }
  int root = buildTree(1, n);
  while (q--) {
    int x, y;
    ignore = scanf("%d %d", &x, &y);
    if (x < 0) {
      x *= -1;
      w[x] = y;
      upd(root, x);
    } else {
      long long S = getSum(root, x, y);
      st_S = (S + 1) / 2;
      int k = get(root, x);
      long long ans = 0;
      ans = getSum(root, x, k - 1) - getSum(root, k + 1, y) + MD;
      ans %= MD;
      ans *= 1LL * a[k];
      ans %= MD;
      ans += getSum2(root, k + 1, y) - getSum2(root, x, k - 1) + MD;
      ans %= MD;
      cout << ans << "\n";
    }
  }
}