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 N = 2E5 + 5;
const int MOD = 1E9 + 7;
int n, m, a[N], w[N], x, y, l, r, mid, p[N], pre[N], ne[N];
long long f[N], tr[2][N], ans, s0, s1, s2, s3, s4;
void add(int k, int x, long long w) {
  for (; x <= n; x += x & -x) {
    tr[k][x] += w;
    if (!k) tr[k][x] %= MOD;
  }
}
long long get(int k, int x) {
  long long sum = 0;
  for (; x; x -= x & -x) {
    sum += tr[k][x];
    if (!k) sum %= MOD;
  }
  return sum;
}
void work(int x, long long k) {
  add(0, x, k * w[x] * a[x] % MOD);
  add(1, x, k * w[x]);
  f[p[x]] += k * w[x];
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    a[i] -= i;
    p[i] = p[i - 1];
    if (i == 1 || a[i] != a[i - 1]) p[i] = p[i - 1] + 1;
    if (p[i] == p[i - 1])
      pre[i] = pre[i - 1];
    else
      pre[i] = i - 1;
  }
  for (int i = n; i; i--)
    if (p[i] == p[i + 1])
      ne[i] = ne[i + 1];
    else
      ne[i] = i + 1;
  for (int i = 1; i <= n; i++) scanf("%d", &w[i]), work(i, 1);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &x, &y);
    if (x < 0) {
      x = -x;
      work(x, -1);
      w[x] = y;
      work(x, 1);
    } else {
      l = x;
      r = y;
      s0 = get(1, x - 1);
      s1 = get(1, y);
      while (l <= r) {
        mid = (l + r) / 2;
        if (pre[mid] >= x)
          s2 = get(1, pre[mid]) - s0;
        else
          s2 = 0;
        if (ne[mid] <= y)
          s3 = s1 - get(1, ne[mid] - 1);
        else
          s3 = 0;
        if (ne[mid] - 1 <= y)
          s4 = get(1, ne[mid] - 1);
        else
          s4 = s1;
        if (pre[mid] >= x)
          s4 -= get(1, pre[mid]);
        else
          s4 -= s0;
        if (abs(s2 - s3) <= s4) break;
        if (s2 < s3)
          l = mid + 1;
        else
          r = mid - 1;
        if (l > r)
          for (;;)
            ;
      }
      ans = get(0, y) - get(0, mid) * 2 + get(0, x - 1) -
            (get(1, y) - get(1, mid) * 2 + get(1, x - 1)) % MOD * a[mid] % MOD;
      ans = (ans % MOD + MOD) % MOD;
      printf("%lld\n", ans);
    }
  }
}