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 MAXN = 220000;
const int P = 1E9 + 7;
int a[MAXN], w[MAXN], s[MAXN];
struct SegmentTree {
  struct Node {
    int l, r;
    long long sum1;
    int sum2;
  } tr[4 * MAXN];
  void pushup(int o) {
    tr[o].sum1 = tr[(o << 1)].sum1 + tr[(o << 1 | 1)].sum1;
    tr[o].sum2 = (tr[(o << 1)].sum2 + tr[(o << 1 | 1)].sum2) % P;
  }
  void build(int o, int l, int r, int* w) {
    tr[o].l = l;
    tr[o].r = r;
    if (l == r) {
      tr[o].sum1 = w[l];
      tr[o].sum2 = 1ll * (l - ::a[l] + P) * w[l] % P;
      return;
    }
    int mid = (l + r) >> 1;
    build((o << 1), l, mid, w);
    build((o << 1 | 1), mid + 1, r, w);
    pushup(o);
  }
  void update(int o, int p, int k) {
    if (tr[o].l == tr[o].r) {
      tr[o].sum1 = k;
      tr[o].sum2 = 1ll * (p - ::a[p] + P) * k % P;
      return;
    }
    int mid = (tr[o].l + tr[o].r) >> 1;
    if (p <= mid)
      update((o << 1), p, k);
    else
      update((o << 1 | 1), p, k);
    pushup(o);
  }
  long long query1(int o, int l, int r) {
    if (tr[o].l > r || tr[o].r < l) return 0;
    if (l <= tr[o].l && tr[o].r <= r) return tr[o].sum1;
    return query1((o << 1), l, r) + query1((o << 1 | 1), l, r);
  }
  int query2(int o, int l, int r) {
    if (tr[o].l > r || tr[o].r < l) return 0;
    if (l <= tr[o].l && tr[o].r <= r) return tr[o].sum2;
    return (query2((o << 1), l, r) + query2((o << 1 | 1), l, r)) % P;
  }
  int find(int o, int l, int r, long long k, long long s = 0) {
    if (tr[o].l == tr[o].r) return tr[o].l;
    int mid = (tr[o].l + tr[o].r) >> 1;
    if (mid >= l && s + tr[(o << 1)].sum1 >= k)
      return find((o << 1), l, r, k, s);
    else
      return find((o << 1 | 1), l, r, k, s + tr[(o << 1)].sum1);
  }
} st;
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
  st.build(1, 1, n, w);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x < 0)
      st.update(1, -x, y);
    else {
      int p = st.find(1, x, y,
                      st.query1(1, 1, x - 1) + (st.query1(1, x, y) + 1) / 2);
      int r1 = (st.query2(1, x, p - 1) +
                st.query1(1, x, p - 1) % P * (a[p] - p + P)) %
               P;
      int r2 = (st.query1(1, p + 1, y) % P * (p - a[p] + P) -
                st.query2(1, p + 1, y) + P) %
               P;
      printf("%d\n", (r1 + r2) % P);
    }
  }
  return 0;
}