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>
using namespace std;
template <typename _T>
inline void _DBG(const char *s, _T x) {
  cerr << s << " = " << x << "\n";
}
template <typename _T, typename... args>
void _DBG(const char *s, _T x, args... a) {
  while (*s != ',') cerr << *s++;
  cerr << " = " << x << ',';
  _DBG(s + 1, a...);
}
const int N = 2e5 + 7;
const int M = 1e9 + 7;
const int MAX = 1 << 18;
struct node {
  long long l, r;
  long long sum, val, pos;
};
void merge(node &nn, node A, node &B) {
  node my;
  my.l = A.l, my.r = B.r;
  my.sum = A.sum + B.sum;
  my.pos = B.pos;
  my.val = (B.val + A.val + A.sum % M * (B.pos - A.pos + A.r - B.r)) % M;
  nn = my;
}
int n, q;
long long A[N], W[N], RA[N], RW[N];
node tree[2 * MAX + 7], rtree[2 * MAX + 7];
long long query1(int l, int r) {
  l += MAX, r += MAX;
  if (l == r) {
    return 0;
  }
  vector<int> vec = {r};
  node res = tree[l];
  while (l / 2 != r / 2) {
    if (l % 2 == 0) merge(res, res, tree[l + 1]);
    if (r % 2 == 1) vec.push_back(r - 1);
    l /= 2, r /= 2;
  }
  while (!vec.empty()) {
    merge(res, res, tree[vec.back()]);
    vec.pop_back();
  }
  return res.val;
}
long long query2(int l, int r) {
  l = n - l + 1, r = n - r + 1;
  swap(l, r);
  l += MAX, r += MAX;
  if (l == r) {
    return 0;
  }
  vector<int> vec = {r};
  node res = rtree[l];
  while (l / 2 != r / 2) {
    if (l % 2 == 0) merge(res, res, rtree[l + 1]);
    if (r % 2 == 1) vec.push_back(r - 1);
    l /= 2, r /= 2;
  }
  while (!vec.empty()) {
    merge(res, res, rtree[vec.back()]);
    vec.pop_back();
  }
  return res.val;
}
long long sum(int l, int r) {
  l += MAX, r += MAX;
  if (l == r) return tree[l].sum;
  long long ans = tree[l].sum + tree[r].sum;
  while (l / 2 != r / 2) {
    if (l % 2 == 0) ans += tree[l + 1].sum;
    if (r % 2 == 1) ans += tree[r - 1].sum;
    l /= 2, r /= 2;
  }
  return ans;
}
long long alls;
long long cost(int i, int l, int r) {
  long long s1 = sum(l, i), s2 = alls - s1;
  return s1 - s2;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
    RA[n - i + 1] = -A[i];
  }
  for (int i = 1; i <= n; i++) {
    cin >> W[i];
    RW[n - i + 1] = W[i];
  }
  for (int i = 1; i <= n; i++) {
    tree[MAX + i].l = tree[MAX + i].r = i;
    tree[MAX + i].sum = W[i], tree[MAX + i].val = 0;
    tree[MAX + i].pos = A[i];
    rtree[MAX + i].l = rtree[MAX + i].r = i;
    rtree[MAX + i].sum = RW[i], rtree[MAX + i].val = 0;
    rtree[MAX + i].pos = RA[i];
  }
  for (int i = MAX; i; i--) {
    merge(tree[i], tree[2 * i], tree[2 * i + 1]);
    merge(rtree[i], rtree[2 * i], rtree[2 * i + 1]);
  }
  while (q--) {
    int x, y;
    cin >> x >> y;
    if (x < 0) {
      x *= -1;
      W[x] = y;
      tree[MAX + x].sum = y;
      for (int e = (x + MAX) / 2; e; e /= 2) {
        merge(tree[e], tree[2 * e], tree[2 * e + 1]);
      }
      RW[n - x + 1] = y;
      rtree[n - x + 1 + MAX].sum = y;
      for (int e = (n - x + 1 + MAX) / 2; e; e /= 2) {
        merge(rtree[e], rtree[2 * e], rtree[2 * e + 1]);
      }
    } else {
      if (x == y) {
        cout << 0 << "\n";
      } else {
        int l = x - 1, r = y;
        alls = sum(x, y);
        while (l + 1 < r) {
          int m = (l + r) / 2;
          if (cost(m, x, y) <= 0) {
            l = m;
          } else {
            r = m;
          }
        }
        l++;
        cout << (query1(x, l) + query2(l, y)) % M << "\n";
      }
    }
  }
  return 0;
}
