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 MOD = 1e9 + 7;
const int N = 200010;
int _w;
void add_mod(int &a, int b) {
  a += b;
  a -= a >= MOD ? MOD : 0;
}
int lowb(int x) { return x & (-x); }
int n, q, a[N], w[N];
struct BIT {
  long long c[N];
  void init() { memset(c, 0, sizeof c); }
  void add(int p, int v) {
    for (; p <= n; p += lowb(p)) c[p] += v;
  }
  long long sum(int p) {
    long long s = 0;
    for (; p; p -= lowb(p)) s += c[p];
    return s;
  }
  long long sum(int l, int r) { return sum(r) - sum(l - 1); }
};
BIT bitw;
struct MODBIT {
  int c[N];
  void init() { memset(c, 0, sizeof c); }
  void add(int p, int v) {
    for (; p <= n; p += lowb(p)) add_mod(c[p], v);
  }
  int sum(int p) {
    int s = 0;
    for (; p; p -= lowb(p)) add_mod(s, c[p]);
    return s;
  }
  int sum(int l, int r) { return (sum(r) - sum(l - 1) + MOD) % MOD; }
};
MODBIT bitaw;
int _calc(int l, int r, int x) {
  x = a[x];
  int sub = bitaw.sum(l, r);
  int plus = int(bitw.sum(l, r) % MOD);
  x = int(1LL * x * plus % MOD);
  return (x - sub + MOD) % MOD;
}
int calc(int l, int r, int x) {
  int t1 = _calc(l, x, x);
  int t2 = _calc(x, r, x);
  t2 = (MOD - t2) % MOD;
  return (t1 + t2) % MOD;
}
void solve(int l, int r) {
  int low = l - 1, high = r;
  while (low < high) {
    int mid = (low + high + 1) / 2;
    long long lsum = bitw.sum(l, mid);
    long long rsum = bitw.sum(mid + 1, r);
    if (lsum <= rsum) {
      low = mid;
    } else {
      high = mid - 1;
    }
  }
  printf("%d\n", calc(l, r, low + 1));
}
int main() {
  _w = scanf("%d%d", &n, &q);
  bitaw.init();
  bitw.init();
  for (int i = 1; i <= n; ++i) {
    _w = scanf("%d", a + i);
    a[i] = (a[i] - i + MOD) % MOD;
  }
  for (int i = 1; i <= n; ++i) {
    _w = scanf("%d", w + i);
    bitw.add(i, w[i]);
    int tmp = int(1LL * a[i] * w[i] % MOD);
    bitaw.add(i, tmp);
  }
  while (q--) {
    int x, y;
    _w = scanf("%d%d", &x, &y);
    if (x < 0) {
      int id = -x;
      int delta = y - w[id];
      w[id] += delta;
      bitw.add(id, delta);
      delta = (delta + MOD) % MOD;
      int tmp = int(1LL * a[id] * delta % MOD);
      bitaw.add(id, tmp);
    } else {
      solve(x, y);
    }
  }
  return 0;
}