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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
int n, q, a[200005];
long long t[524333], w[200005];
struct T {
  long long t[200005];
  void add(int x, long long k) {
    while (x <= n) t[x] += k, x += (x & -x);
  }
  long long sum(int x) {
    long long s = 0;
    while (x) s += t[x], x -= (x & -x);
    return s;
  }
  long long get(int l, int r) { return sum(r) - sum(l - 1); }
} f1, f2;
int fnd(int k, long long o) {
  if (k > 262143) return k - 262143;
  int lc = k << 1;
  if (t[lc] >= o) return fnd(lc, o);
  return fnd(lc ^ 1, o - t[lc]);
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i), a[i] = a[i] - i + 1;
  for (int i = 1; i <= n; ++i) {
    scanf("%I64d", w + i);
    t[262143 + i] = w[i];
    f1.add(i, w[i]);
    f2.add(i, w[i] * 1ll * a[i] % 1000000007ll);
  }
  for (int i = 262143; i; --i) t[i] = t[i << 1] + t[(i << 1) ^ 1];
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    if (l < 0) {
      l = -l;
      f1.add(l, -w[l]);
      f2.add(l, -(w[l] * 1ll * a[l] % 1000000007ll));
      w[l] = r;
      f1.add(l, w[l]);
      f2.add(l, w[l] * 1ll * a[l] % 1000000007ll);
      int k = l + 262143;
      t[k] = r;
      do {
        k >>= 1;
        t[k] = t[k << 1] + t[(k << 1) ^ 1];
      } while (k > 1);
      continue;
    }
    long long L = f1.sum(l - 1), R = f1.sum(r), O = R - L;
    int k = fnd(1, L + (O + 1ll) / 2ll);
    long long ans =
        a[k] * 1ll * (f1.get(l, k - 1) % 1000000007ll) % 1000000007ll -
        f2.get(l, k - 1) % 1000000007ll + f2.get(k + 1, r) % 1000000007ll -
        a[k] * 1ll * (f1.get(k + 1, r) % 1000000007ll) % 1000000007ll;
    ans = (ans % 1000000007ll + 1000000007ll) % 1000000007ll;
    printf("%I64d\n", ans);
  }
  return 0;
}
