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(3)
#pragma GCC optimize(2)
using namespace std;
const long long mod = 1e9 + 7;
struct segt {
  long long wt, d;
  segt operator+(segt other) const {
    segt ret;
    ret.wt = this->wt + other.wt;
    ret.d = (this->d + other.d) % mod;
    return ret;
  }
};
segt tree[524292];
long long x[524292], w[524292];
int n, q;
void build(int l, int r, int index) {
  if (l > r) return;
  if (l == r) {
    tree[index].wt = w[l];
    tree[index].d = w[l] * (x[l] - l);
    return;
  }
  build(l, ((l + r) >> 1), (index << 1));
  build(((l + r) >> 1) + 1, r, (index << 1 | 1));
  tree[index] = tree[(index << 1)] + tree[(index << 1 | 1)];
}
void update(int pt, int l, int r, int index, long long val) {
  if (l > r || pt < l || pt > r) return;
  if (l == r) {
    w[pt] = val;
    tree[index].wt = w[pt];
    tree[index].d = w[pt] * (x[pt] - l);
    return;
  }
  if (pt <= ((l + r) >> 1)) {
    update(pt, l, ((l + r) >> 1), (index << 1), val);
  } else {
    update(pt, ((l + r) >> 1) + 1, r, (index << 1 | 1), val);
  }
  tree[index] = tree[(index << 1)] + tree[(index << 1 | 1)];
}
segt query(int start, int end, int l, int r, int index) {
  if (start > end || l > r || start > r || l > end) return tree[0];
  if (start <= l && r <= end) return tree[index];
  if (end <= ((l + r) >> 1)) {
    return query(start, end, l, ((l + r) >> 1), (index << 1));
  }
  if (start >= ((l + r) >> 1) + 1) {
    return query(start, end, ((l + r) >> 1) + 1, r, (index << 1 | 1));
  }
  return query(start, end, l, ((l + r) >> 1), (index << 1)) +
         query(start, end, ((l + r) >> 1) + 1, r, (index << 1 | 1));
}
int main() {
  int i;
  scanf("%d%d", &n, &q);
  for (i = 1; i <= n; ++i) {
    scanf("%I64d", &x[i]);
  }
  for (i = 1; i <= n; ++i) {
    scanf("%I64d", &w[i]);
  }
  build(1, n, 1);
  while (q-- > 0) {
    long long L, R;
    scanf("%I64d%I64d", &L, &R);
    if (L < 0) {
      update(-L, 1, n, 1, R);
    } else {
      if (L == R) {
        printf("0\n");
        continue;
      }
      int low = L, high = R;
      int pivot = low;
      auto W = query(L, R, 1, n, 1).wt;
      while (low <= high) {
        int mid = (low + high) >> 1;
        auto ret1 = query(L, mid, 1, n, 1);
        if (ret1.wt >= W - ret1.wt) {
          pivot = mid;
          high = mid - 1;
        } else {
          low = mid + 1;
        }
      }
      auto ret1 = query(L, pivot, 1, n, 1);
      auto ret2 = query(pivot + 1, R, 1, n, 1);
      long long cost1 = (x[pivot] - pivot) * (ret1.wt % mod) - ret1.d;
      long long cost2 = ret2.d + (ret2.wt % mod) * (pivot - x[pivot]);
      long long ans = (cost1 + cost2) % mod;
      if (ans < 0) ans += mod;
      printf("%I64d\n", ans);
    }
  }
  return 0;
}
