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 INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const int mod = 1e9 + 7;
const int N = 2e5 + 5;
struct segT {
  int n;
  long long *seg;
  bool f;
  segT() {}
  segT(const vector<int> &v, bool f) : f(f) {
    n = v.size();
    seg = new long long[4 * N + 1];
    build(1, 0, n - 1, v);
  }
  void build(int r, int i, int j, const vector<int> &v) {
    if (i == j) {
      seg[r] = v[i];
    } else {
      int mid = (i + j) >> 1;
      build(2 * r, i, mid, v);
      build(2 * r + 1, mid + 1, j, v);
      seg[r] = seg[2 * r] + seg[2 * r + 1];
      if (f and seg[r] >= mod) seg[r] -= mod;
    }
  }
  int a, b;
  long long query(int r, int i, int j, long long val) {
    if (j < a or i > b) return 0;
    if (i >= a and j <= b) {
      if (val != -1) seg[r] = val;
      return seg[r];
    } else {
      int mid = (i + j) >> 1;
      long long L = query(2 * r, i, mid, val);
      long long R = query(2 * r + 1, mid + 1, j, val);
      if (f) {
        seg[r] = seg[2 * r] + seg[2 * r + 1];
        if (seg[r] >= mod) seg[r] -= mod;
        if (L + R >= mod) return L + R - mod;
        return L + R;
      } else {
        seg[r] = seg[2 * r] + seg[2 * r + 1];
        return L + R;
      }
    }
  }
  long long query(int l, int r) {
    a = l;
    b = r;
    return query(1, 0, n - 1, -1);
  }
  void update(int p, long long val) {
    a = b = p;
    query(1, 0, n - 1, val);
  }
} segw, segaw, segiw;
int n, q;
int a[N], w[N];
int get(int a, int b, long long l, long long beg) {
  long long ret = ((segw.query(a, b) % mod) * beg) % mod;
  ret = (ret + segiw.query(a, b));
  ret = (ret - segaw.query(a, b));
  ret = (ret - (segw.query(a, b) % mod) * l) % mod;
  return ret;
}
int go(int l, int r) {
  l--;
  r--;
  long long sum = segw.query(l, r);
  int bot = l, top = r;
  int center = 0;
  while (bot <= top) {
    int mid = (bot + top) >> 1;
    if (segw.query(l, mid) * 2LL >= sum) {
      center = mid;
      top = mid - 1;
    } else {
      bot = mid + 1;
    }
  }
  int beg = a[center] - (center - l);
  int L = get(l, center, l, beg);
  int R = -get(center + 1, r, l, beg);
  int ret = (L + R) % mod;
  if (ret < 0) ret += mod;
  return ret;
}
int main(void) {
  vector<int> vw, vaw, viw;
  scanf("%d %d", &n, &q);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  for (int i = 0; i < n; i++) {
    scanf("%d", w + i);
    vw.push_back(w[i]);
    vaw.push_back((1LL * w[i] * a[i]) % mod);
    viw.push_back((1LL * w[i] * i) % mod);
  }
  segw = segT(vw, false);
  segaw = segT(vaw, true);
  segiw = segT(viw, true);
  while (q--) {
    int x, y;
    scanf("%d %d", &x, &y);
    if (x > 0) {
      printf("%d\n", go(x, y));
    } else {
      x *= -1;
      x--;
      segw.update(x, y);
      segaw.update(x, (1LL * a[x] * y) % mod);
      segiw.update(x, (1LL * y * x) % mod);
    }
  }
  return 0;
}