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 MAX_N = 2e5 + 200, mod = 1e9 + 7;
int n, q, ai[MAX_N], bi[MAX_N], wi[MAX_N];
long long sum[MAX_N << 2], diff[MAX_N << 2];
void pushup(int p) {
  sum[p] = sum[(p << 1)] + sum[((p << 1) | 1)],
  diff[p] = (diff[(p << 1)] + diff[((p << 1) | 1)]) % mod;
}
void build(int l, int r, int p) {
  if (l == r) {
    sum[p] = wi[l], diff[p] = 1LL * wi[l] * bi[l];
    return;
  }
  build(l, ((l + r) >> 1), (p << 1)),
      build(((l + r) >> 1) + 1, r, ((p << 1) | 1));
  pushup(p);
}
void update(int qx, int l, int r, int p) {
  if (l == r) {
    sum[p] = wi[l], diff[p] = 1LL * wi[l] * bi[l];
    return;
  }
  if (qx <= ((l + r) >> 1))
    update(qx, l, ((l + r) >> 1), (p << 1));
  else
    update(qx, ((l + r) >> 1) + 1, r, ((p << 1) | 1));
  pushup(p);
}
long long query(int ql, int qr, int l, int r, int p, long long *nodes,
                bool opt = false) {
  if (ql > qr) return 0;
  if (ql <= l && r <= qr) return nodes[p];
  long long ret = 0;
  if (ql <= ((l + r) >> 1))
    ret += query(ql, qr, l, ((l + r) >> 1), (p << 1), nodes);
  if (((l + r) >> 1) < qr)
    ret += query(ql, qr, ((l + r) >> 1) + 1, r, ((p << 1) | 1), nodes);
  if (opt) ret %= mod;
  return ret;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &ai[i]), bi[i] = ai[i] - i;
  for (int i = 1; i <= n; i++) scanf("%d", &wi[i]);
  build(1, n, 1);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    if (l < 0)
      l = -l, wi[l] = r, update(l, 1, n, 1);
    else {
      int lb = l, rb = r, res = l;
      while (lb <= rb) {
        int mid = (lb + rb) >> 1;
        if (query(l, mid, 1, n, 1, sum) >= query(mid + 1, r, 1, n, 1, sum))
          res = mid, rb = mid - 1;
        else
          lb = mid + 1;
      }
      long long part1 =
          (1LL * query(l, res - 1, 1, n, 1, sum, true) * (ai[res] - res) % mod +
           mod - query(l, res - 1, 1, n, 1, diff, true)) %
          mod;
      long long part2 = (1LL * query(res + 1, r, 1, n, 1, sum, true) *
                             (res + mod - ai[res]) % mod +
                         query(res + 1, r, 1, n, 1, diff, true)) %
                        mod;
      printf("%lld\n", (0LL + part1 + part2) % mod);
    }
  }
  return 0;
}