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]. 
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
const long long LLINF = 0x3f3f3f3f3f3f3f3f;
const int MAX_N = (int)3e5 + 17;
const int mod = (int)1e9 + 7;
int N, M, S, T, K;
int a[MAX_N], w[MAX_N];
struct Fenwick_0 {
  long long bt[MAX_N];
  void init() {
    for (int i = 0; i <= N; i++) bt[i] = 0;
  }
  void add(int idx, int w) {
    for (; idx <= N; idx += idx & -idx) bt[idx] += w;
  }
  long long sum(int idx) {
    long long ans = 0;
    for (; idx > 0; idx -= idx & -idx) ans += bt[idx];
    return ans;
  }
} W;
struct Fenwick_1 {
  int bt[MAX_N];
  void init() {
    for (int i = 0; i <= N; i++) bt[i] = 0;
  }
  void add(int idx, int w) {
    for (; idx <= N; idx += idx & -idx) (bt[idx] += w) %= mod;
  }
  int sum(int idx) {
    int ans = 0;
    for (; idx > 0; idx -= idx & -idx) (ans += bt[idx]) %= mod;
    return ans;
  }
} F;
int main() {
  while (~scanf("%d %d", &N, &M)) {
    W.init();
    F.init();
    for (int i = 1; i <= N; i++) {
      scanf("%d", &a[i]);
      a[i] -= i;
    }
    for (int i = 1; i <= N; i++) {
      scanf("%d", &w[i]);
      W.add(i, w[i]);
      F.add(i, 1LL * a[i] * w[i] % mod);
    }
    while (M--) {
      int x, y;
      scanf("%d %d", &x, &y);
      if (x < 0) {
        x *= -1;
        W.add(x, y - w[x]);
        F.add(x, 1LL * (y - w[x]) * a[x] % mod);
        w[x] = y;
      } else {
        long long sl = W.sum(x - 1), sr = W.sum(y);
        sr = (sr - sl + 1) / 2;
        int l = x, r = y;
        while (l <= r) {
          int mid = (l + r) >> 1;
          if (W.sum(mid) - sl >= sr)
            r = mid - 1;
          else
            l = mid + 1;
        }
        long long ans = (F.sum(y) - F.sum(l)) -
                        (W.sum(y) - W.sum(l)) % mod * 1LL * a[l] % mod -
                        ((F.sum(l) - F.sum(x - 1)) -
                         (W.sum(l) - W.sum(x - 1)) % mod * 1LL * a[l] % mod);
        printf("%lld\n", (ans % mod + mod) % mod);
      }
    }
  }
  return 0;
}
